What is the answer to this limit what is the limit value of the power series:
$$ \lim_{x\rightarrow +\infty} \sum_{k=1}^\infty (-1)^k \frac{x^k}{k^{k-m}}$$
where $m>1$.

For $\lim_{x\rightarrow +\infty} \sum_{k=1}^\infty (-1)^k \frac{x^k}{k^{k-m}}$,
the ratio of consecutive terms is
$\begin{align}
\frac{x^{k+1}}{(k+1)^{k+1-m}}\big/\frac{x^k}{k^{k-m}}
&=\frac{x k^{k-m}}{(k+1)^{k+1-m}}\\
&=\frac{x }{k(1+1/k)^{k+1-m}}\\
&=\frac{x }{k(1+1/k)^{k+1}(1+1/k)^{-m}}\\
&\approx\frac{x }{ke(1+1/k)^{-m}}\\
&=\frac{x (1+1/k)^{m}}{ke}\\
\end{align}
$
For large $k$ and fixed $m$,
$(1+1/k)^{m} \approx 1+m/k$,
so the ratio is about
$x/(ke)$,
so the sum is like
 $\lim_{x\rightarrow +\infty} \sum_{k=1}^\infty (-1)^k \big(\frac{x}{ek}\big)^k$.
Note: I feel somewhat uncomfortable
about this conclusion, but I will continue anyway.
Interesting how the $m$ goes away (unless I made a mistake, which is not unknown).
In the answer pointed to by Mhenni Benghorbal,
it is shown that this limit is $-1$,
so it seems that this is the limit of this sum also.
