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Ratio Test Methodology

  • Thread starter Rapier
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  • #1
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Homework Statement



Determine whether Ʃ(1→∞) n^2/e^n converges or diverges.

Homework Equations



L = lim (n→∞) abs [a_n+1/a_n]

The Attempt at a Solution



The prof was out of town so left us a "self-study" task. We're looking at the Ratio Test and I want to see if my methodology is correct.

a_n+1 = (n+1)^2/e^n+1 = n^2 + 2n + 1/e^n+1
a_n = n^2/e^n

L = lim (n→∞) abs [a_n+1/a_n]
L = lim (n→∞) abs [n^2 + 2n + 1/e^(n+1) / n^2/e^n]
L = lim (n→∞) abs [n^2 + 2n + 1 / (e*n^2)]
L = lim (n→∞) abs [(n^2/n^2 + 2n / n^2 + 1/n^2)/(e*n^2/n^2)] <--- ?
L = lim (n→∞) abs [(1 + 1/n + 1/n^2) / e]
L = 1/e

The spot I'm not sure about is dividing the terms by the highest term in the denominator. When I did that for the next problem (n/2^n) I got L = .5 and it can't possibly converge to .5 because the first two terms are .5.

Does the existence of the limit simply mean convergence and not what the limit is converging to? I seem to remember something like that, but can't recall for sure.

Thanks for the help.
 

Answers and Replies

  • #2
33,306
4,998

Homework Statement



Determine whether Ʃ(1→∞) n^2/e^n converges or diverges.

Homework Equations



L = lim (n→∞) abs [a_n+1/a_n]

The Attempt at a Solution



The prof was out of town so left us a "self-study" task. We're looking at the Ratio Test and I want to see if my methodology is correct.

a_n+1 = (n+1)^2/e^n+1 = n^2 + 2n + 1/e^n+1
Arrrghh! You need parentheses! The right side is n2 + 2n + 1/en + 1. I hope that you recognize that this is not the same as (n2 + 2n + 1)/(en + 1) or even (n2 + 2n + 1)/en + 1. I'm sure that's not what you intended.
a_n = n^2/e^n

L = lim (n→∞) abs [a_n+1/a_n]
L = lim (n→∞) abs [n^2 + 2n + 1/e^(n+1) / n^2/e^n]
L = lim (n→∞) abs [n^2 + 2n + 1 / (e*n^2)]
L = lim (n→∞) abs [(n^2/n^2 + 2n / n^2 + 1/n^2)/(e*n^2/n^2)] <--- ?
L = lim (n→∞) abs [(1 + 1/n + 1/n^2) / e]
L = 1/e
You don't need all of this. Just look at the ratio, an+1/an, and take the limit. You don't need absolute values, since all of the terms in your series are already positive.

The limit of the ratio above is 1/e, so what does the Ratio Test tell you about the related series? This should be in your textbook.
The spot I'm not sure about is dividing the terms by the highest term in the denominator. When I did that for the next problem (n/2^n) I got L = .5 and it can't possibly converge to .5 because the first two terms are .5.

Does the existence of the limit simply mean convergence and not what the limit is converging to? I seem to remember something like that, but can't recall for sure.
The Ratio Test can tell you whether a given series converges, but it doesn't say what the series converges to.
 
  • #3
87
0
Arrrghh! You need parentheses! The right side is n2 + 2n + 1/en + 1. I hope that you recognize that this is not the same as (n2 + 2n + 1)/(en + 1) or even (n2 + 2n + 1)/en + 1. I'm sure that's not what you intended.
You don't need all of this. Just look at the ratio, an+1/an, and take the limit. You don't need absolute values, since all of the terms in your series are already positive.

The limit of the ratio above is 1/e, so what does the Ratio Test tell you about the related series? This should be in your textbook.
The Ratio Test can tell you whether a given series converges, but it doesn't say what the series converges to.
Yeah, I probably should have parenthised that a little better. Sorry about that. Thanks for the help!
 

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