Limit of a summation 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!
 A: 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. $\frac{1}{1-2}=-1$).
A: 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.
A: 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 
$$
\lim_{n \rightarrow \infty} \frac{x^n}{n+a} = \infty
$$
