Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
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.

share|cite|improve this answer
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
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. $\frac{1}{1-2}=-1$).

share|cite|improve this answer
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:… – 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 $$ \lim_{n \rightarrow \infty} \frac{x^n}{n+a} = \infty $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.