Ratio test | Series | AP Calculus BC | Khan Academy
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- čas přidán 3. 09. 2014
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The ratio test is a most useful test for series convergence. It caries over intuition from geometric series to more general series. Learn more about it here.
AP Calculus BC on Khan Academy: Learn AP Calculus BC - everything from AP Calculus AB plus a few extra goodies, such as Taylor series, to prepare you for the AP Test
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"this isn't some voodoo" actually made me laugh out loud. Love you Sal!
sal: this isn't voodoo
everyone: actually it is
Very useful! Thanks!
Thank u so much....Now i can literally Visiulize the Seriea🤩🤩♥️
6:45 "yes we can" sounded like Obama lol 😭
Hahaha
Thanxs yall are genius
tank you very much sir,
you make it easy to understand.
Shoutout to Ms. Swish! Loved your PE classes. Keep on surfin' and catchin' those mavericks!
Thank You 🤩 🤩 🤩
I love you man
Can you make videos on English like vocabulary and grammar
Shoutout to Mr. Achille’s class!
Word.
What about harmonic series? It's a divergent series. 1+ 1/2 + 1/3 + 1/4 .....
how many of you watching this in college
paying how much a year for the teachers to show me this and TAs to help more
You can't prove that n^10/n! doesn't diverge with the divergence test as stated in 2:40 because lim an = 0 so it can still diverge ou converge. Or am I wrong?
Correct, divergence test is inconclusive!
Yes, divergence test has only two possible ending: the series diverges or the test is inconclusive.
Divergence test cannot be used to prove the convergence of a series.
You are right. For this video on the official website, there was a comment block that appeared right when he said it that wrote "The divergence test cannot be used for this case".
what happen if the limit of (n+1)th term over nth term doesn't approach to any value? We simply cannot use Ratio Test, can we?
Zerobyte Johnsons That's right. Using the ratio test requires that the limit actually exists and is a finite number.
thx dude, just wanna clarify.
I'm afraid I still cannot tell what it happening when you say that n is going to infinity. If n becomes infinity in both the denominator and the numerator of a fraction, isn't it just going to converge at 1 regardless?
Working with infinity and limits is weird, but basically the way it works here is the denominator in this case “approaches infinity faster” than the number are because it is a higher power. For example lim of n/n^2 approaches 0 because the denominator has a higher power. You can solve this by dividing the top and bottom by the highest power, so it would be come 1/n. And the lim as n approaches infinity of 1/n is one over a really big number which approaches 0
L + ratio
Do you have videos that teach like other subjects a and languages?
They do. All the videos are on khanacademy.com. I suggest following the playlists.
Thank you so much!!! martyspandex
+deanna boneham yes
how do we decide when to write n=5, n=1 or n=0? I am kind of confused :S
it's given to you, becomes part of the question. Most will start at n=1 or n=0 for simplicity in finding a few initial terms
1:40 I feel that if abs(r) = 1, the series STILL converges to the first term
I realize this is from 2 years ago so apologies for bringing it up now as its probably not relevant for you anymore, but I thought it was still a question that deserved an answer- more for the sake of anyone who's reading these comments now with similar questions.
If the ratio is 1, the sequence of terms would most certainly all be the same as the first term (since we're just multiplying by 1 each time), but what we mean when we talk about the convergence or divergence of a series is the *sum* of all of those terms. If the terms aren't approaching 0 as n --> infinity, then that sum can't be a finite number since you'll be continually adding new numbers to it (this is the idea behind the Divergence Test). In our case with r = 1, you're adding the same term over and over forever, so of course that infinite sum could only be written as being "equal" to infinity (in other words, it is a divergent series since the sum does not get closer and closer to equaling some finite number).
Basically, what you've pointed out is that the *sequence of terms* "converges" to the first term as n --> infinity, but this is different from the series converging to a finite number (which, as the Divergence Test will show, can only occur if that sequence of terms goes to 0 as n --> infinity.
@UCM74r8Kt9bGF4XPA61iFfVw I'm so glad it was helpful for you! It's definitely one of those things that I was incredibly happy to figure out for myself, I remember it being a big deal for me when that concept started to click. Are you in Calc 2 by any chance? Just curious, since that's where I was first introduced to this stuff (last semester for me)
I guess =1 can be a different case...but definately that tends to infinity thts why Divergent suits the more
@@Litt13F00t 🙏♥️
une swag
Can’t watch this man at less then 2x speed lol
Tell me k hmy kesy pta chly ga k is question ma kon sa test use hona hy?
I thought the ratio of a geometric series was An/An-1?
no, it is a1/(1-r)
and r is the common ration between terms
Just the same....Succeeding term divided by Preceding Term
In every Khan academy video he verbally repeats any text he is writing several times and it drives me fucking crazy
I don't even care about how many times he repeats himself, his teaching is absolutely astonishing
it helps to stuck infos to our brains. its annoying as shit and i remember all of them because i was angry at that moment.
sexy voice'
You might want to consider avoiding the hand as the pointer - it is severely distracting!
Prof. Leonard > Khan academy
Prof. Leonardo + Khan good mix
Or just learn from both.