Hey everyone! Thanks for watching. I'm aware that at 1:38 my explanation was rushed and not very rigorous - sometimes when videos are too long I try and skip over important things like that! A more convincing argument may be to find the difference between the terms n+1 and n instead.
I'm not clear why you need to go through the process of solving for S_infinity. Consider π/(3sqrt3) > 3/(3sqrt3) = 1/sqrt3 > 1/3 Consider S_1 = 1/(1^2+1x1+1^2) = 1/3 Since S_1 = 1/3, we know that not all S_n can be > π/(3sqrt3) > 1/3 Therefore, A
Since the condition has to hold for all n's and either A or B has to be true. So all you have to do is substitute n = 1, you get S1 = 1/3. now, we can prove that pi/(3sqrt(3)) > 1/3 (pretty easy to do this). And hence the correct option is A. This question is to be done in under 1 minute in the exam. You cannot waste that much time on it.
If those are the only two options, we can just take the value of S1 which is 1/3, this is already less than pi/(3*root(3)), so it's clear the second option can't be true. But of course, in the real exam there are four options, so it will be much harder. The solution is very elegant but one never has the time to construct such a solution in a exam like JEE :P
Loved the part where you turned the series into an integral using the basic integral definition....but I think the part where you told the function S (n) was increasing with n was not intuitive...because although the number of terms increase the individual terms are become smaller ...anyway that was my only issue otherwise it's pretty awesome video...
Thanks so glad you enjoyed! Yeah I probably over explained that bit - just meant n approaching infinity would provide the maximum we're looking for. Thanks for the comment!
@@OscgrMaths Yes not just a question of adding a summand s(n+1,n+1) when going from S_n to S_(n+1) also to compare s(k,n) with s(k,n+1) for all k =< n. Given that n is in the numerator & denominator for the summand s(k,n) I think there is a step missing to show that S_infy is indeed the maximum.
@@andrewbuchanan5342 I thought the explanation was correct. As all terms are clearly positive, however small, S_m > S_n whenever m>n, simply because you are adding more positive terms when going from n to m, and so the max is indeed S_infy
@@silver6054It's true that you have more positive terms in the S_m case than the S_n case when m > n, but the terms themselves are smaller due to the n^2 component in the denominator. You'd have to confirm that the terms being smaller isn't enough to offset there being more of them
@@_Heb_ yes, you are right, totally missed the key point that S_m and S_n contain different terms, (i.e. the terms of S_n are not a subset of the terms of S_m). So yes, definitely need an argument that S_infy is a maximum
beautiful use of the riemann sum! some of my friends are preparing for JEE in thr next few years and they have practice problems like this all the time. its a really challenging exam esp when you realise you can only dedicate 3ish minutes per question. anyways superb solution!
Bro, I have cleared JEE and am currently studying at one of the top colleges in India, IIT KGP, basically the MIT or Stanford of India. This question is way too easy, obviously. There are some questions that are easy and some that are hard, but if you want to taste the real flavor of JEE, try the 2016 or 2022 papers, especially old JEE papers like those from 1995 to easy 2000s . Those questions were real gems.
Ohh boy this is not even the hardest limit as a sum integration question lol Jee main (simpler version of jee advanced) has had harder limit as a sum question than this
@@OscgrMaths czcams.com/users/live8erjxlTASgo?feature=shared This question is considered to be the toughest one, it uses the same concept of limit of a sum by integration which you talked about in this video
By the way I do not get the argument for Sn to be monotonically increasing. Sure, with higher n, we are adding more positive terms, but we are also adding *different* terms, probably smaller terms because we have a n^2 in the denominator
This is definitely not the hardest jee adv problem... The only tricky part is knowing Sn < S∞ and after that it's a really standard problem which most of the students can solve in 1 min
@@ionex18 Fair enough - do you have any harder ones I could cover on the channel 😁? I found this one from looking around on stack exchange for hardest maths questions from the JEE but if there's even harder I'd love to see! Thanks for the comment.
@@OscgrMaths Oh yea sure... You can solve jee adv 2016 paper 1 question 43... Though it's similar question on Riemann summation... Many people regard it as the toughest question.... Or you can check out other questions in 2016 coz it was the toughest jee adv paper...
How did you deduce that S_n < S_∞ ? From what I see, the Riemann sum of 1/n. f(k/n), k running from 1 to n gives a bigger value than the actual area under the curve y=f(x), because it is the Riemann upper sum. Am I missing anything?? Edit: I got it. Since f(x) is a decreasing function in the interval [0,1], so the Riemann sum actually gives us the Lower Riemann Sum.
This problem looks familiar... I was gonna say that n=1 thing but I got beaten to it lol, nonetheless this is a crazy solution imagine having to see that Riemann sum in the middle of the exam Also congrats on 2k
@@OscgrMaths totally depends on your exposure. As an Indian student, we solve lots of problems pertaining to reinmann sums. So, literally the first thing, I would observe when i see this question is, hey that is the reimann sum without the limit, and the options clearly indicate that some integration is probably involved. To set up the inequality, I drew a rough graph of 1/(1+x+x^2). The total area under the curve is the value of Sn as n approaches inf i.e. rectangular strips are infinitely small. Sn with n being finite represents an approximate area under the curve. When you start k from 1 to n, the summation represents the area with rectangles of base length 1/n. It can be graphically observed that this area is less then the actual area under the curve, similar logic for second part
I am not sure how you readily conclude S_n is an increasing sequence of numbers. if you take the difference of S_{n+1} and S_n, you do get an additional term at the top (at n+1) but the difference of the other terms must be dealt with
If the inequality is supposed to hold for all values of n, just substitute n=1. Then, since pi/3sqrt(3) ~ 22/21sqrt(3) > 1/sqrt(3) > 1/3 = S1, the answer is A.
@@OscgrMaths Yeah. It's certainly not a legit solve, but it cheeses the multiple-choice format nicely. (I understand these poor students are ALSO severely time-crunched on this exam.)
@@user-en5vj6vr2u Haha yes. I did not miss that, but there is always a small possibility that the formula doesn't work for border cases. Generally in such cases the question puts a limit on the value of two; but since you are supposed to solve this analytically they might intentionally skip it. So it is safer to put n=2, but in most cases n=1 may also work; and in this problem it should work (I guess).
I used the integral test right from the get go and was able to solve it with a tighter bound I think. However, the integral was a bit of a pain and led to a less neat solution.
@@OscgrMaths FYI, JEE is "Joint Entrance Examination" and the question must be probably from JEE Advanced, the one that is used to determine the entry of the participants into IIT's and IISER's. JEE Main is a slightly easier exam and is for entry into NITs, IIIT's and some other such institutes (i.e., those excluding IITs and such institutes) and also serve as an elimination round for the participants to attend JEE Advanced. Still, thanks for covering the question.
@@samrubenabraham6979slight correction: IISERS closed the JEE channel recently making the only way of entry to any IISER through their own entrance exam, but iisc still accepts jee scores.
I remember solving this Q and more like this one back in 11th grade, we are actually taught riemann sum as a part of problem solving techniques for questions involving series and sequences, this is actually among the easier ones
@@vaanivijay65522022, IIT Delhi, do PYQs, give paper during the same time as the exams, focus on a strong theoretical foundation and doing conceptually tough questions, adv pyqs are a must then you can move on to other books
Congratulations on another fantastic video! Your style is lucid and engaging, and you've clearly thought deeply about the exact order to present each step. This is the third video I've seen of yours - I've liked and subscribed :)
didnt watch the video yet cuz i wanted to do it myself but doesnt the sum become a riemann integral and it would be integral from 0 to1 of 1/(1+x+x^2) which is kinda easy to solve
Your explanation of why S_n is increasing is shakey. Yes you keep adding up more and more parts, but those parts become smaller and smaller. You could compute S_n+1 - S_n and match the k term from S_n+1 with the k+1 term from S_n and see that indeed the difference of those two terms is positive.
@@red0guy I see my mistake now. Each term in the sum depends on n, but so does the upper limit of the sum. I agree with you, this is a little trickier than it first appears.
Can you please tell me in which year this question came i jee?? And I wanna make sure that are you taking about the jee advance for high school students?!
well well well..... if it has to hold for any integer n, it also has to work for n=1, i plug that and I get 1/3,which is less than that number, so i solved it lol
But even if they are smaller it will still make a slight increase each time, meaning the maximum value must be as n goes to infinity since there are no more terms left to add.
@@OscgrMaths This reasoning is incorrect. Here's a simple counterexample: The exact same sum but switch the bounds from 1->n to 0->n-1. Here you have to identify the summation as a right Riemann Sum, and because the function f is continuous and decreasing on the interval [0; 1], then the sequence is approaching the integral from below, thus S_nn-1, you can see that the summation is a left Riemann Sum, and since f is continuous and decreasing on [0; 1], then the sequence is approaching the integral from above, that is, S_inf
@@LinaWainwright That's what the challenge question at the end is all about - it's the exact same sum but starting at a different place. Sorry for the explanation in the video and thanks for typing this all out!
Can i get why cant i just get n=1 to be a way to solve, since S1=1/3 and we now that √3 1 then S1 < π/3√3 and since its just an A, B question then then we can deduce that for all n and say that A is the correct answer
I have solve this question while practicing some previous year and I did not even thought it was hard😅 I tried another way it states that the the inequality of SN so we can have n equal to any number so I put n equal to one and I got one third which is smaller but I could be wrong
This is so simple. sum(n, k=1) n/(n^2+kn+k^2) => sum(n, k=1) n/((n^3-k^3)(n-k)) => sum(n, k=1) n(n-k)/(n^3-k^3) now that it is simplified... when k reaches n, n-k = 0 so we can ignore the last term(except when taken as 1*0/0 instead take as lim x->0 1*0/x = 0/x = 0 n=1 => sum(1, k=1) 1(1-1)/(1^3-1^3) use lim x->0 1*0/x = 0/x = 0 0 2(2-1)/(2^3-1^3) + 2(2-2)/(2^3-2^3) the second term or the nth term is why we use lim x->0 1*0/x = 0/x = 0 => 2/73 is enough
Not to be that guy but i don't this was too hard. Riemann sums seem the obvious way to go. Also, from the problem set up, you can guess that the integral is equal to pi/sqrt(27) and skip half the work
@@OscgrMaths I don't know what S_n < pi/3sqrt3 means because S_n is not a number but a sequence. Does it mean instead the limit as n goes to infinity of S_n < pi / 3sqrt3, and does it always mean that when you have a sequence in an inequality or equation?
@@MathematicFanatic Whatever you choose n as is the upper limit of your sum, so you're looking for an inequality that holds for all n in the natural numbers. Hope this helps!
@@OscgrMaths By this interpretation, the choices are (S_n < pi/3sqrt3 for all values of n) and (S_n > pi/3sqrt3 for all values of n). This makes the problem extremely easy because I can just check that S_1 = 1/3 so I know that the second option is false so it must be the first option.
this is not the hardest problem in jee but nonethanless this is an interesting question(and one of the challenging ones to solve in those restricting time bound situations)
What do you even mean by that😂😂 the guy who have solved this sum called it harder which means It's hard for him Indians can solve this question within minutes ever heard about JEE advanced exam?😂😂 Sundar ceo of Google gave that exam crack it and got into iit(dream college of many jee aspirants)and now you know he is ceo of Google it's second hardest exam of the entire world just after gaokao from china the difficulty of that paper is so much that scoring 40 percent in the exam will get you in iit 😂😂 I bet you can't even score 1percent in that exam you will be in negative😂😅
@@joaquingomez9226 @SalmonForYourLuck I am joking chill guys, I literally subbed to his channel since months and i saw all his vids i like his content, this is my first comment and i thought it would be funny to write it 😮💨but i guess in internet people take everything seriously
@@joaquingomez9226 @SalmonForYourLuck I am joking chill guys, I literally subbed to his channel since months and i saw all his vids and i like his content. This is my first comment and i thought it would be funny to write it 😮💨but i guess in internet people take everything seriously
Hey everyone! Thanks for watching. I'm aware that at 1:38 my explanation was rushed and not very rigorous - sometimes when videos are too long I try and skip over important things like that! A more convincing argument may be to find the difference between the terms n+1 and n instead.
This is hard until you realize you only have a 50% chance to get it wrong
There are 4 options tho, and there is -2 marks if you get the answers wrong
Try INMO problems. You will get them on aops, it is the toughest high school Math exam in India.
I'm not clear why you need to go through the process of solving for S_infinity.
Consider π/(3sqrt3) > 3/(3sqrt3) = 1/sqrt3 > 1/3
Consider S_1 = 1/(1^2+1x1+1^2) = 1/3
Since S_1 = 1/3, we know that not all S_n can be > π/(3sqrt3) > 1/3
Therefore, A
Since the condition has to hold for all n's and either A or B has to be true. So all you have to do is substitute n = 1, you get S1 = 1/3. now, we can prove that pi/(3sqrt(3)) > 1/3 (pretty easy to do this). And hence the correct option is A.
This question is to be done in under 1 minute in the exam. You cannot waste that much time on it.
This is exactly what I was going to do
If those are the only two options, we can just take the value of S1 which is 1/3, this is already less than pi/(3*root(3)), so it's clear the second option can't be true. But of course, in the real exam there are four options, so it will be much harder. The solution is very elegant but one never has the time to construct such a solution in a exam like JEE :P
Loved the part where you turned the series into an integral using the basic integral definition....but I think the part where you told the function S (n) was increasing with n was not intuitive...because although the number of terms increase the individual terms are become smaller ...anyway that was my only issue otherwise it's pretty awesome video...
Thanks so glad you enjoyed! Yeah I probably over explained that bit - just meant n approaching infinity would provide the maximum we're looking for. Thanks for the comment!
@@OscgrMaths Yes not just a question of adding a summand s(n+1,n+1) when going from S_n to S_(n+1) also to compare s(k,n) with s(k,n+1) for all k =< n. Given that n is in the numerator & denominator for the summand s(k,n) I think there is a step missing to show that S_infy is indeed the maximum.
@@andrewbuchanan5342 I thought the explanation was correct. As all terms are clearly positive, however small, S_m > S_n whenever m>n, simply because you are adding more positive terms when going from n to m, and so the max is indeed S_infy
@@silver6054It's true that you have more positive terms in the S_m case than the S_n case when m > n, but the terms themselves are smaller due to the n^2 component in the denominator. You'd have to confirm that the terms being smaller isn't enough to offset there being more of them
@@_Heb_ yes, you are right, totally missed the key point that S_m and S_n contain different terms, (i.e. the terms of S_n are not a subset of the terms of S_m). So yes, definitely need an argument that S_infy is a maximum
Love u bro from India
Thank you so much!
beautiful use of the riemann sum! some of my friends are preparing for JEE in thr next few years and they have practice problems like this all the time. its a really challenging exam esp when you realise you can only dedicate 3ish minutes per question. anyways superb solution!
Thanks! Glad you enjoyed.
@@OscgrMaths come on man i am preparing for jee and this was not the hardest , again in india these are common ideas
Bro, I have cleared JEE and am currently studying at one of the top colleges in India, IIT KGP, basically the MIT or Stanford of India. This question is way too easy, obviously. There are some questions that are easy and some that are hard, but if you want to taste the real flavor of JEE, try the 2016 or 2022 papers, especially old JEE papers like those from 1995 to easy 2000s . Those questions were real gems.
Ohh boy this is not even the hardest limit as a sum integration question lol
Jee main (simpler version of jee advanced) has had harder limit as a sum question than this
What is the hardest JEE question you've ever seen? I might do that one next!
@@OscgrMaths czcams.com/users/live8erjxlTASgo?feature=shared
This question is considered to be the toughest one, it uses the same concept of limit of a sum by integration which you talked about in this video
Loved it, fun problem and crystal clear explanations. So glad I found your channel!
@@forgottenwishes7775 Thanks so much! Really glad you enjoyed.
that ain't no exam that's projecteuler 😭
TRY OUT THE INTEGRATION PROBLEM FROM JEE ADV 2008 IT WAS A NIGHTMARE FOR MANY AS FAR AS I HAVE HEARD FROM OTHERS :)))
@@studyingpuppets8916 Okay thank you!!
@@OscgrMaths welcome and also one from jee adv 2010 paper 1 on integration.
By the way I do not get the argument for Sn to be monotonically increasing. Sure, with higher n, we are adding more positive terms, but we are also adding *different* terms, probably smaller terms because we have a n^2 in the denominator
I was just solving this pyq and it was discussed in our class today 😂😂😂
❤thanks buddy for accepting my request❤ you remembered me? ☺️
@@Aman-b5k Of course! I'm going to do even more soon hopefully. Thanks for such a good request.
remember solving this in class, it was so much fun!
Glad you enjoyed!
This is definitely not the hardest jee adv problem...
The only tricky part is knowing Sn < S∞ and after that it's a really standard problem which most of the students can solve in 1 min
@@ionex18 Fair enough - do you have any harder ones I could cover on the channel 😁? I found this one from looking around on stack exchange for hardest maths questions from the JEE but if there's even harder I'd love to see! Thanks for the comment.
JEE Adv 2019 probability question!
Jee adv famous limit question (you can search on yt u,will easily find it @@OscgrMaths
@@Rudrubro11 Thanks!
@@OscgrMaths Oh yea sure...
You can solve jee adv 2016 paper 1 question 43...
Though it's similar question on Riemann summation... Many people regard it as the toughest question....
Or you can check out other questions in 2016 coz it was the toughest jee adv paper...
How did you deduce that S_n < S_∞ ?
From what I see, the Riemann sum of 1/n. f(k/n), k running from 1 to n gives a bigger value than the actual area under the curve y=f(x), because it is the Riemann upper sum. Am I missing anything??
Edit: I got it. Since f(x) is a decreasing function in the interval [0,1], so the Riemann sum actually gives us the Lower Riemann Sum.
Observe
n>=k
=》n^2+nk+k^2>= 3k^2
=》1/n^2+nk+k^2
🎉easier ,this are training q for those who prepare for jee adv
This problem looks familiar...
I was gonna say that n=1 thing but I got beaten to it lol, nonetheless this is a crazy solution imagine having to see that Riemann sum in the middle of the exam
Also congrats on 2k
@@gregoriousmaths266 Thank you! Yeah Riemann sum part is totally crazy.
@@OscgrMaths totally depends on your exposure. As an Indian student, we solve lots of problems pertaining to reinmann sums. So, literally the first thing, I would observe when i see this question is, hey that is the reimann sum without the limit, and the options clearly indicate that some integration is probably involved.
To set up the inequality, I drew a rough graph of 1/(1+x+x^2). The total area under the curve is the value of Sn as n approaches inf i.e. rectangular strips are infinitely small.
Sn with n being finite represents an approximate area under the curve. When you start k from 1 to n, the summation represents the area with rectangles of base length 1/n. It can be graphically observed that this area is less then the actual area under the curve, similar logic for second part
I am not sure how you readily conclude S_n is an increasing sequence of numbers. if you take the difference of S_{n+1} and S_n, you do get an additional term at the top (at n+1) but the difference of the other terms must be dealt with
Yeah there was a bit more work behind the scenes here.. sorry for brushing over it!
Great content; do not clickbait though.
Jee is nowhere the toughest exam, and this is definitely no the toughest question on jee.
If u want the hardest problem, check the jee adv 2016 limits problem and jee adv 2020 matrices problem.
@@arjishchowdhury1898 Okay great thanks!
If the inequality is supposed to hold for all values of n, just substitute n=1. Then, since pi/3sqrt(3) ~ 22/21sqrt(3) > 1/sqrt(3) > 1/3 = S1, the answer is A.
@@alexandersanchez9138 This is a nice strategy.
@@OscgrMaths Yeah. It's certainly not a legit solve, but it cheeses the multiple-choice format nicely. (I understand these poor students are ALSO severely time-crunched on this exam.)
@@alexandersanchez9138 Definitely a huge time pressure!!!
Yeah when I first saw the problem I thought of exactly that. Either way, enjoyed the video 😂
bro isnt that Sn>S∞ because the whole sum is going to zero as n=∞
and if i put the values of n as 1 and 2
i got S1>S2 = 3/4>9/7
This is not the hardest question
Nobody sees this any hard
The real hard questions are completely different
I put n=2, and summed two terms to check. That's how you should do it in the exam.
It's not about how you do it in exam. It's about how your intuition works your perspective of looking the problem look how beautifully I explained.
@@shivangsingh5834 Thank you professor. I know what math is about. Have you cracked JEE by any chance?
Step your game up buddy. It said n may be equal to 1, so set it equal to 1!
@@user-en5vj6vr2u Haha yes. I did not miss that, but there is always a small possibility that the formula doesn't work for border cases. Generally in such cases the question puts a limit on the value of two; but since you are supposed to solve this analytically they might intentionally skip it. So it is safer to put n=2, but in most cases n=1 may also work; and in this problem it should work (I guess).
I cant stop watching your videos! 😂 so interesting, when I grow up i want to be a mathematician
By the way students take this on a computer in the actual exam
Thats great!! So glad you're enjoying.
n=1 simple
had trouble understanding riemann sums
walked out understood everything
Wow I'm really glad!
I used the integral test right from the get go and was able to solve it with a tighter bound I think. However, the integral was a bit of a pain and led to a less neat solution.
great work man i loved the vid and the explanation was great
@@O_of_N Thanks so much! Glad you enjoyed.
I compared it with base problem with inequality
@@JOSHUVASRINATH Nice!
@@OscgrMaths FYI, JEE is "Joint Entrance Examination" and the question must be probably from JEE Advanced, the one that is used to determine the entry of the participants into IIT's and IISER's. JEE Main is a slightly easier exam and is for entry into NITs, IIIT's and some other such institutes (i.e., those excluding IITs and such institutes) and also serve as an elimination round for the participants to attend JEE Advanced. Still, thanks for covering the question.
@@samrubenabraham6979slight correction: IISERS closed the JEE channel recently making the only way of entry to any IISER through their own entrance exam, but iisc still accepts jee scores.
I remember solving this Q and more like this one back in 11th grade, we are actually taught riemann sum as a part of problem solving techniques for questions involving series and sequences, this is actually among the easier ones
Damn bro u in IIT now???
@@vaanivijay6552 fortunately, yep
@@udaykaranwal1662 wow congratulations! Which year did u write IIT JEE? Which iit r u in now? Any tips for us aspirants bro?
@@vaanivijay65522022, IIT Delhi, do PYQs, give paper during the same time as the exams, focus on a strong theoretical foundation and doing conceptually tough questions, adv pyqs are a must then you can move on to other books
Congratulations on another fantastic video! Your style is lucid and engaging, and you've clearly thought deeply about the exact order to present each step. This is the third video I've seen of yours - I've liked and subscribed :)
Thanks so much! That really means a lot.
Great video
It is not the hardest of india also not the hardest problem😂😂😂😂
Then what is the toughest competitive exam of India?
Keep in mind that u have to solve this in 5 min only
I know, that's why the JEE is so tough!
Brother please do more JEE questions we Indians love you😊😊
@@sohampine7304 Will do!! Thanks for the comment.
didnt watch the video yet cuz i wanted to do it myself but doesnt the sum become a riemann integral and it would be integral from 0 to1 of 1/(1+x+x^2) which is kinda easy to solve
Yes!! That's the satisfying part, nice spot.
@@OscgrMaths thanks you love this sums please cover more problems from jee i never heard of it
@@bingchilling4717 Will do!
Which university are you from bro
Your explanation of why S_n is increasing is shakey. Yes you keep adding up more and more parts, but those parts become smaller and smaller. You could compute S_n+1 - S_n and match the k term from S_n+1 with the k+1 term from S_n and see that indeed the difference of those two terms is positive.
@@red0guy I see my mistake now. Each term in the sum depends on n, but so does the upper limit of the sum. I agree with you, this is a little trickier than it first appears.
Can you please tell me in which year this question came i jee?? And I wanna make sure that are you taking about the jee advance for high school students?!
This one is from 2008! Hope that helps.
This is preety easy and also i think this is not the toughest jee question
Would be nice if more hard questions like these are made as your videos :)
Thank you!
well well well..... if it has to hold for any integer n, it also has to work for n=1, i plug that and I get 1/3,which is less than that number, so i solved it lol
not proved obviously but since it has to be either a or b, it must be a by this reasoning
Nice! Very smart...
@@OscgrMaths lazyness can bring you far sometimes
why does S_n < S_inf ? as n approaches infinity number of terms increases but they are also smaller
But even if they are smaller it will still make a slight increase each time, meaning the maximum value must be as n goes to infinity since there are no more terms left to add.
@@OscgrMaths ah I see, thank you!
It's my bad for skipping over that in the video, there is a way to be more rigorous about it. Thanks for raising the question!
@@OscgrMaths This reasoning is incorrect. Here's a simple counterexample: The exact same sum but switch the bounds from 1->n to 0->n-1.
Here you have to identify the summation as a right Riemann Sum, and because the function f is continuous and decreasing on the interval [0; 1], then the sequence is approaching the integral from below, thus S_nn-1, you can see that the summation is a left Riemann Sum, and since f is continuous and decreasing on [0; 1], then the sequence is approaching the integral from above, that is, S_inf
@@LinaWainwright That's what the challenge question at the end is all about - it's the exact same sum but starting at a different place. Sorry for the explanation in the video and thanks for typing this all out!
The question maybe a lil tricky but as questions on same approach have been asked before so it becomes easy to know the solution
making jee more fun
Can i get why cant i just get n=1 to be a way to solve, since S1=1/3 and we now that √3 1 then
S1 < π/3√3 and since its just an A, B question then then we can deduce that for all n and say that A is the correct answer
🙂 thanks but how can we solve this under 3mins
He is explaoning no one solve in 3 min
Shortcut is substituting n=1, given that the inequality has to work for all n.
I have solve this question while practicing some previous year and I did not even thought it was hard😅 I tried another way it states that the the inequality of SN so we can have n equal to any number so I put n equal to one and I got one third which is smaller but I could be wrong
@@aadeshr4307 That's a nice shortcut!
As an 11 th greader i tried to understand this but i am not able to hope in couple of month i will 😅
Btw i am preparing for iit jee
Nice explanation bro - never seen anything like this but still managed to follow pretty well. Keep it up boss
@@user-hr4vr6oj3v Thank you! Really glad you enjoyed, thanks for the comment.
Cant it be dn using telescoping?
@@cliche_ayush I'm not quite sure what you mean but it's definitely worth a shot - if you solve it with a new method let me know and I'd love to see!
I assume you could just re index the challenge question and tackle it a similar way to the original?
@@Er4serOP Yeah exactly... that's a good place to start!
Can this be solved without knowing riemann's sum?
Riemann sums is how I did it but if you manage to find another approach I'd love to know!!
Bro is a genius
I wouldn't say that!! But thanks for the comment.
This is so simple.
sum(n, k=1) n/(n^2+kn+k^2)
=> sum(n, k=1) n/((n^3-k^3)(n-k))
=> sum(n, k=1) n(n-k)/(n^3-k^3)
now that it is simplified...
when k reaches n, n-k = 0 so we can ignore the last term(except when taken as 1*0/0 instead take as lim x->0 1*0/x = 0/x = 0
n=1
=> sum(1, k=1) 1(1-1)/(1^3-1^3)
use lim x->0 1*0/x = 0/x = 0
0 2(2-1)/(2^3-1^3) + 2(2-2)/(2^3-2^3)
the second term or the nth term is why we use lim x->0 1*0/x = 0/x = 0
=> 2/73 is enough
Not to be that guy but i don't this was too hard. Riemann sums seem the obvious way to go. Also, from the problem set up, you can guess that the integral is equal to pi/sqrt(27) and skip half the work
It becomes hard when you consider the limited time and the education level of those who take it.
Does the statement S_n < number mean that the limit of S_n is less than the number?
Hey what do you mean with limit of S_n? Thanks for the question!
@@OscgrMaths I don't know what S_n < pi/3sqrt3 means because S_n is not a number but a sequence. Does it mean instead the limit as n goes to infinity of S_n < pi / 3sqrt3, and does it always mean that when you have a sequence in an inequality or equation?
@@MathematicFanaticApparently in this problem Sn is a sum not a sequence, I also got confused by the notation
@@MathematicFanatic Whatever you choose n as is the upper limit of your sum, so you're looking for an inequality that holds for all n in the natural numbers. Hope this helps!
@@OscgrMaths By this interpretation, the choices are (S_n < pi/3sqrt3 for all values of n) and (S_n > pi/3sqrt3 for all values of n). This makes the problem extremely easy because I can just check that S_1 = 1/3 so I know that the second option is false so it must be the first option.
How old are you?
This was sick, I've never seen a problem like this before
Thanks! I thought it was quite unique too, which is why I wanted to share.
this is not the hardest problem in jee but nonethanless this is an interesting question(and one of the challenging ones to solve in those restricting time bound situations)
@@PooshanHalder Yeah definitely! I found it by looking up hardest JEE question but I'm sure there will be others that are even harder!!
@@OscgrMaths There is a question from JEE ADV 2016 paper You'll love to solve that
It might difficult for Indians, but not the most difficult one throughout the world. The difficulty level is clearly mediocre.
What do you even mean by that😂😂 the guy who have solved this sum called it harder which means It's hard for him Indians can solve this question within minutes ever heard about JEE advanced exam?😂😂 Sundar ceo of Google gave that exam crack it and got into iit(dream college of many jee aspirants)and now you know he is ceo of Google it's second hardest exam of the entire world just after gaokao from china the difficulty of that paper is so much that scoring 40 percent in the exam will get you in iit 😂😂 I bet you can't even score 1percent in that exam you will be in negative😂😅
Skill issue
How old are you
@@SalmonForYourLuck mentally or chronologically?
@@anas.aldadi This answer makes it clear, thanks!
@@joaquingomez9226 @SalmonForYourLuck I am joking chill guys, I literally subbed to his channel since months and i saw all his vids i like his content, this is my first comment and i thought it would be funny to write it 😮💨but i guess in internet people take everything seriously
@@joaquingomez9226 @SalmonForYourLuck I am joking chill guys, I literally subbed to his channel since months and i saw all his vids and i like his content. This is my first comment and i thought it would be funny to write it 😮💨but i guess in internet people take everything seriously