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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!

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The series diverges for every value of $x$ such that $|x|\ge1$ hence the limit when $x\to\infty$ is not defined. –  Did Jun 21 '11 at 5:10

3 Answers 3

up vote 5 down vote accepted

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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Though I clearly specify that $x\to\infty$. Wouldn't WolframAlpha put out some message before making that assumption? –  kevmo314 Jun 21 '11 at 5:15
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If you have $A(B)$ WolframAlpha first evaluates $B=C$ and then $A(C)$, in your case $B$ is the sum and $A$ the limit. Note that it sets $\sum _{n=1}^{\infty } \frac{x^n}{n+1}=\frac{-x-\log (1-x)}{x}$ which is only true if $|x|<1$ –  Listing Jun 21 '11 at 5:17
    
Ah, that explanation makes sense. Thank you very much. –  kevmo314 Jun 21 '11 at 5:22
    
No problem, I agree that WolframAlpha can easily confuse you if you don't see how it evaluates expressions. –  Listing Jun 21 '11 at 5:31

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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But how does this explain what WolframAlpha is doing? Is it making a mistake? –  kevmo314 Jun 21 '11 at 5:11
    
@kevm0314: Could you provide a link / post the commands you are using in Wolfram Alpha? –  Zev Chonoles Jun 21 '11 at 5:13
    
Yes, added it to the original post if this link doesn't work: wolframalpha.com/input/… –  kevmo314 Jun 21 '11 at 5:14
    
@kevm0314: Thanks for the link. I've added to my answer. –  Zev Chonoles Jun 21 '11 at 5:19
    
Ah I wish I could accept multiple answers. Both of your answers were incredibly helpful. Thanks so much! –  kevmo314 Jun 21 '11 at 5:29

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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