Sir, in 2nd question you take the divergent test , which says that if the limit n--》infinite (An) is not= 0 then it diverges.. but if the value is 0 then there is no conclusion that its converges or not ... so we need other testing method for it ..i was waiting for it. But sir just said which is converges.. how ?
@@muhsinm4257 The Divergence Test is only for series and not sequences. Don't be confused with convergence of sequences and series. The sequence a_n=1/n is convergent because its limit as n goes to infinity exists--it is 0; however, the series, sum of 1/n from n=1 to infinity, is divergent. That is well known as harmonic series.
When I am asked to test whether the series is convergent or divergent I proceed as an+1 - an and i check if it's less than 0 and finally put limit n-♾️ in an
Hello. Squareroot of n^3 is like the dominant term in the denominator. Similar strategy with finding limits at infinity of rational functions where we divide all the terms in the numerator and denominator by the highest power of x in the denominator.
Don't be confused with convergence of sequences and series. A sequence is convergent if the limit exists, that is, the limit is a real number. Foe example, the sequence a_n=1/n is convergent because its limit as n goes to infinity exists--it is 0; however, the series, sum of 1/n from n=1 to infinity, is divergent. That is well known as harmonic series.
How can you say if limit of nth term is zero when n tends to infinity...you are wrong here sir... Because we can not tell the convergence or divergence of it for sure...we need further test...sir plzz cheak it.. For example if An=1/n in this case your method fails..
Don't be confused with convergence of sequences and series. The sequence a_n=1/n is convergent because its limit as n goes to infinity exists--it is 0; however, the series, sum of 1/n from n=1 to infinity, is divergent. That is well known as harmonic series.
Divide the numerator and denominator by n then you can easily determine the limit. And from there you can determine whether it is convergent or divergent.
Hello Jessa. Tama na kapag ang numerator ay papuntang infinity at ang denominator ay papuntang 1, yung fraction ay papunpang infinity/1 o infinity. Pero kapag pagpilitin ang numerator at denominator, magiging 1/infinity at yang ang papuntang 0.😊
Don't be confused with convergence of sequences and series. The sequence a_n=1/n is convergent because its limit as n goes to infinity exists--it is 0; however, the series, sum of 1/n from n=1 to infinity, is divergent. That is well known as harmonic series.
This is similar to problem 1. Divide all the terms in your numerator and denominator by n^2, simplify the little fractions, and from there you can easily determine the number to which it converges.
Typo in solution of 1: the difference in the numerator must be addition
Sir, in 2nd question you take the divergent test , which says that if the limit n--》infinite (An) is not= 0 then it diverges.. but if the value is 0 then there is no conclusion that its converges or not ... so we need other testing method for it ..i was waiting for it. But sir just said which is converges.. how ?
@@muhsinm4257 The Divergence Test is only for series and not sequences. Don't be confused with convergence of sequences and series. The sequence a_n=1/n is convergent because its limit as n goes to infinity exists--it is 0; however, the series, sum of 1/n from n=1 to infinity, is divergent. That is well known as harmonic series.
I like how everything is just so conventional...there are no difficult problems or easy problems...we just have problems...so calming
Crisp and Concise, helped a lot :)
Please for the first exercise why did you change the sign of 3 in the numerator?from +3 to -3
Woww!! Helped me understand the limits of sequence better.
My new favorite math channel
I am confused (question 3) how are we having √n after dividing n^2 in √n^3
THANKYOU SO MUCH SIR..
RESPECT FROM PAKISTAN
It's really helpful video... Thanku so much sir..
Glad it helps!
Good work sir it really very easy understand thank u for this video
Glad it helped
great selection of examples. thank you
Thanks for explanation, do you have video for proving that as well?
When I am asked to test whether the series is convergent or divergent
I proceed as an+1 - an and i check if it's less than 0 and finally put limit n-♾️ in an
Perfect 👍🏼👍🏼it's resourceful
Dear Professor K.O.MATH,
In question 7, how could we know that we divide both denominator and numerator for 3^n? Thank you very much.
Because it is the dominant term in the denominator, similar to the technique when finding limits at infinity of rational functions.
Very helpful.
Very helpful sir,make more videos for bsc 5th semester
Glad to help. More videos coming soon🙂
Hello sir,
(1+1/n)^-5 is convergent but how to find its boundness
Please sir explain this my exams are coming
Perfect 👍.. it's so understandable
Impressive taught 👏 👌
Very helpful video,thankuu sir🙏
Helps me a lot. Thank u so much sir
Glad it helped!
Sir in 1st problem, if we take n^4 common answer is 1. Then it will be convergent sir. Please clarify my doubt sir.
Could you clarify your question? If you meant that if the leading term in the denominator is n^4 then the answer is 1, then you are correct.
what is the use and aplication of these pls someone tel me >.< ahh!!!!!!!!!!!
Thank you very much sir i watch just 1st one and i am able to solve my problems ❤️
Glad to help!
Quand utilise-t-on la règle de L'Hubetal ?? Est à 0/0 et aussi à &/&
Thanks Sir
hi sir, may i know on question 3, why did you multiply the numerator and denominator with 1/squareroot n3?
Hello. Squareroot of n^3 is like the dominant term in the denominator. Similar strategy with finding limits at infinity of rational functions where we divide all the terms in the numerator and denominator by the highest power of x in the denominator.
Why n=0
in question 5, the limit is not equal to zero. shouldnt it be divergent?
Don't be confused with convergence of sequences and series. A sequence is convergent if the limit exists, that is, the limit is a real number. Foe example, the sequence a_n=1/n is convergent because its limit as n goes to infinity exists--it is 0; however, the series, sum of 1/n from n=1 to infinity, is divergent. That is well known as harmonic series.
I wanna to know about eigenvalues
How can you say if limit of nth term is zero when n tends to infinity...you are wrong here sir... Because we can not tell the convergence or divergence of it for sure...we need further test...sir plzz cheak it.. For example if An=1/n in this case your method fails..
Don't be confused with convergence of sequences and series. The sequence a_n=1/n is convergent because its limit as n goes to infinity exists--it is 0; however, the series, sum of 1/n from n=1 to infinity, is divergent. That is well known as harmonic series.
Dear Professor K.O.MATH,
How could I find the limit of this sequence an = 3^n / 4^n
Thank you so much.
This can be written as (3/4)^n and because |3/4|
Please help me in this question sir. Un=2n/5n-3. Determine if it is convergent or divergent.thankyou
Divide the numerator and denominator by n then you can easily determine the limit. And from there you can determine whether it is convergent or divergent.
God bless you sir🙏
Thank you and same to you sir!
Dear Professor,
How could I find the (n-3)! please?
(n-3)!=(n-3)(n-4)...(2)(1)
❤❤❤❤. But you didn't time the exercises😭
😂
ありがと先生
19:00
Hello po pwese magtanong a number 3 po infinity divided by 1 infinity po ba answer? Hindi po 0?
Hello Jessa. Tama na kapag ang numerator ay papuntang infinity at ang denominator ay papuntang 1, yung fraction ay papunpang infinity/1 o infinity. Pero kapag pagpilitin ang numerator at denominator, magiging 1/infinity at yang ang papuntang 0.😊
How do you say an is convergent surely when an = 0 because it is necessary but not sufficient....please correct it will confuse many student..
Don't be confused with convergence of sequences and series. The sequence a_n=1/n is convergent because its limit as n goes to infinity exists--it is 0; however, the series, sum of 1/n from n=1 to infinity, is divergent. That is well known as harmonic series.
@@KOMATH Thanks for explaination.
بالنسبه للمثال الأول لازم نقسم على أكبر اس n4
Solution for number 7 is wrong, right? Shouldn't it be 0?
Hello Benjamin! Which part is wrong? It seems that there is nothing wrong in the solution.
2nd question m 1+1/x kaise aya koi btaega
In 8 que. It should be divergent. Plz someone explain..
It is convergent as the limit exists. May I know why you think it is divergent?
Very helpful 😊
Glad it was helpful!
Sir kindly solve this problem \{(2n ^ 2 + 1)/(2n ^ 2 - 1)\}The sequence
converges to
This is similar to problem 1. Divide all the terms in your numerator and denominator by n^2, simplify the little fractions, and from there you can easily determine the number to which it converges.
@@KOMATH thank you very much sir
Very helpful thank youuuu💕
found u hahaha
Phd hahahaha
Can you tell me convergence for An=3^n/n^2
You can tell the convergence by applying ratio test.
God send
((n+2)^25/2^n+3)(2n/(n+5)^25) Please help me answer this sir..
You can show that (n+2)^25/(n+5)^25 goes to 1 and 2n/(2^n+3) goes to 0 so that sequence approach 0.
What if n -> 25
At what time in the video?
The great vedio
Thanks!
@@KOMATH most welcom😊
thankyuo kan bira dabali ammas
Good
Thanks
👍🏿
Nice
Please for the first exercise why did you change the sign of 3 in the numerator?from +3 to -3
Sorry that is a mistake as written in the pinned comment.😊