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 $m>1$.

Here is a start. Using the integral representation derived in a previous problem and the identity

$$ (xD)^m=\sum_{r=0}^{m}\begin{Bmatrix} m\\r \end{Bmatrix}x^r D^r, $$

where $D=\frac{d}{dx}$ and $ \begin{Bmatrix} m\\r \end{Bmatrix} $ are Stirling numbers of the second kind. 
$$\sum_{k=1}^\infty (-1)^k \frac{x^k}{k^{k}}  = - \int_{0}^{1} x \cdot u^{x u} \, du .$$
$$ \implies (xD)^m\sum_{k=1}^\infty (-1)^k \frac{x^k}{k^{k}}=\sum_{k=1}^\infty (-1)^k k^m\frac{x^k}{k^{k}}=-(xD)^m \int_{0}^{1} x \cdot u^{x u} \, du.$$

$$ \implies \sum_{k=1}^\infty (-1)^k k^m\frac{x^k}{k^{k}}=\sum_{r=0}^{m}\begin{Bmatrix} m\\r \end{Bmatrix} \int_{0}^{1} f(x,u) du , $$

where $f(x,u)$ is a function in $x$ and $u$, see the note. Now, it is your job to try analyzing these integrals and see what the limit is.
Notes: 
1) We used the identity

$$ D^r xe^{ax}= {a}^{r-1}{{\rm e}^{ax}} \left( ax+r \right)  .$$

For finding the $n$th derivative, see section 6. 
2) 

$$ f(x, u) = { {m{x}^{2\,m} \left( \ln  \left( u \right)\right) ^{m-1}{u}^{ux+m
-1}}{ }}+{x}^{2\,m+1} \left( u\ln  \left( u
 \right)  \right) ^{m}{u}^{ux}.$$

