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I'm having trouble evaluating this limit:

$ \lim_{x\to\infty} \sum_{n=1}^\infty \frac{x^n}{(n+a)} $

My intuition and initial attempts at making sense of it say that it diverges, and so do a few of my friends, but WolframAlpha says it equals $-\frac{1}{a}$ (if you plug in some values for $a$) and the intermediate steps for those are pretty useless.

For reference: WolframAlpha's evaluation

Can anyone at least point me in the right direction on how to evaluate this limit?

Thanks in advance!

  • 1
    The series diverges for every value of $x$ such that $|x|\ge1$ hence the limit when $x\to\infty$ is not defined.2011-06-21

3 Answers 3

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Another way to see that it diverges for $x > 1$ is seeing that the general term does not go to $0$ as $n \rightarrow \infty$ because $ \lim_{n \rightarrow \infty} \frac{x^n}{n+a} = \infty $

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It diverges. As soon as $x\geq 1$, we have $\sum_{n=1}^\infty\frac{x^n}{n+a}\geq\sum_{n=1}^\infty\frac{1}{n+a}\geq\sum_{n=\lceil a\rceil+1}^\infty\frac{1}{n},$ which differs from the (divergent) harmonic series by a finite amount. Thus the series is divergent for all $x\geq1$. I believe that Listing's explanation is correct, i.e. WolframAlpha is finding an analytic continuation of the sum, then taking the limit. This is the same reason that $\sum_{n=0}^{\infty}2^n$ diverges, but $\sum_{n=0}^\infty x^n=\frac{1}{1-x}$ and $\frac{1}{1-x}$ is defined for all $x\neq 1$ (e.g. $\frac{1}{1-2}=-1$).

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    @kevm0314: No problem, glad to help!2011-06-21
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The limit is increasing, so it is especially bigger than what you get for $x=1$ where it already diverges.

$\lim_{x\to\infty} \sum_{n=1}^\infty \frac{x^n}{(n+a)}>\sum_{n=1}^\infty \frac{1}{(n+a)}$

Wolfram alpha assumes first that $x$ is chosen in a manner that the sum converges and afterwards calculates the limit for $x$ towards infinity which is not what you want.

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    No problem, I agree that WolframAlpha can easily confuse you if you don't see how it evaluates expressions.2011-06-21