<s>[INST] Asymptotics of $\Sigma x^n/n^p$ I am  looking for asymptotics of 
$$\sum_{n=1}^N \dfrac{x^n}{n^p}$$
in terms of $N$, where $x>1$ and  $p$ is a  positive number.
So I am looking for a $O(f(N))$ type result.  For example what is the approximate value of 
$\sum_{n=1}^N \dfrac{5^n}{n^2}$ for large $N$. One possibility is probably something like $\frac{5^N}{N^2} (a+\frac{b}{N}+\frac{c}{N^2}+\cdots)$.
 [/INST] Just playing around to see what happens.
If $x > 1$
then
$\begin{array}\\
\sum_{n=1}^N \dfrac{x^n}{n^p}
&=x^N\sum_{n=1}^N \dfrac{x^{n-N}}{n^p}\\
&=x^N\sum_{n=0}^{N-1} \dfrac{x^{-n}}{(N-n)^p}\\
&=\dfrac{x^N}{N^p}\sum_{n=0}^{N-1} \dfrac{x^{-n}}{(1-n/N)^p}\\
&=\dfrac{x^N}{N^p}\sum_{n=0}^{N-1} x^{-n}(1-n/N)^{-p}\\
&=\dfrac{x^N}{N^p}\left(1+\dfrac1{x(1-1/N)^p}+\dfrac1{x^2(1-2/N)^p}+...\right)\\
\end{array}
$
To see how (or if)
this converges,
look at the ratio of
consecutive terms.
$\begin{array}\\
\dfrac{x^{-n-1}(1-(n+1)/N)^{-p}}{x^{-n}(1-n/N)^{-p}}
&=\dfrac{(N-n)^p}{x(N-n-1)^{p}}\\
&=\dfrac{1}{x(1-1/(N-n))^{p}}\\
&\approx\dfrac{1}{x(1-p/(N-n))}
\qquad\text{for small } n\\
&\approx\dfrac{1}{x}(1+p/(N-n))\\
&\approx\dfrac{1}{x}(1+p/(N(1-n/N))\\
&\approx\dfrac{1}{x}(1+(p/N)(1+n/N))\\
&=\dfrac{1}{x}(1+p/N+pn/N^2)\\
&\approx\dfrac{1}{x}(1+p/N)\\
\end{array}
$
This is $1$ when
$x(1-1/(N-n))^{p}
=1$
or
$1-1/(N-n)
=x^{-1/p}
$
or
$N-n
=\dfrac{1}{1-x^{-1/p}}
$.
Using the approximate ratio of
$\approx\dfrac{1+p/N}{x}
$,
the approximate expansion is
$\dfrac{x^N}{N^p}\left(1+\dfrac{1+p/N}{x}+\dfrac{(1+p/N)^2}{x^2}+...\right)
$.
That's as far as I can take it now.
Hope this helps.
</s>