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 $m>1$.
 A: My solution is also similar. Let $\left\{ {a \atop b} \right\}$ be the Stirling number of the second kind, as Mhenni Benghorbal pointed out. Then from the identity
$$ x^{n} = \sum_{j=0}^{n} \left\{ {n \atop j} \right\} (x)_{j}, $$
where $(x)_{j} = x (x-1) \cdots (x-j+1)$ is the failing factorial, we obtain
$$ x^{n} = \sum_{j=0}^{n} \left\{ {n+1 \atop j+1} \right\} (x-1)_{j}. $$
This allow us to write 
\begin{align*}
\frac{1}{k^{k-m}}
& = \frac{k^{m}}{k^k} = \sum_{j=0}^{m} \left\{ {m+1 \atop j+1} \right\} \frac{1}{(k-1-j)!} \frac{\Gamma(k)}{k^{k}} \\
&= \sum_{j=0}^{m} \left\{ {m+1 \atop j+1} \right\} \frac{1}{(k-1-j)!} \int_{0}^{\infty} t^{k-1}e^{-kt} \, dt.
\end{align*}
Here, we use the convention that $\frac{1}{j!}$ is zero when $j < 0$, which is compatible with the fact that the reciprocal of the Gamma function is an entire function with zeros at nonpositive integers. Then
\begin{align*}
\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^{k-m}} x^{k}
&= \sum_{k=1}^{\infty} (-1)^{k-1} x^{k} \sum_{j=0}^{m} \left\{ {m+1 \atop j+1} \right\} \frac{1}{(k-1-j)!} \int_{0}^{\infty} t^{k-1}e^{-kt} \, dt \\
&= x \sum_{j=0}^{m} \left\{ {m+1 \atop j+1} \right\} \int_{0}^{\infty} \left( \sum_{k=1}^{\infty}  \frac{(-1)^{k-1}}{(k-1-j)!} \left( x t e^{-t} \right)^{k-1} \right) \, e^{-t} \, dt \\
&= x \sum_{j=0}^{m} \left\{ {m+1 \atop j+1} \right\} \int_{0}^{\infty} \left( \sum_{k=0}^{\infty}  \frac{(-1)^{k+j}}{k!} \left( x t e^{-t} \right)^{k+j} \right) \, e^{-t} \, dt \\
&= x \sum_{j=0}^{m} (-1)^{j} \left\{ {m+1 \atop j+1} \right\} \int_{0}^{\infty} \left( x t e^{-t} \right)^{j} \exp \left( x t e^{-t} \right) \, e^{-t} \, dt \\
&= \sum_{j=0}^{m} (-1)^{j} \left\{ {m+1 \atop j+1} \right\} x^{j+1} \int_{0}^{\infty} t^{j} e^{-tj} \exp \left( x t e^{-t} \right) \, e^{-t} \, dt.
\end{align*}
Here, I considered the negation of your sum in order for typographical simplicity. Anyway, with the substitution $u = e^{-t}$, we obtain
\begin{align*}
\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^{k-m}} x^{k}
&= \sum_{j=0}^{m} (-1)^{j} \left\{ {m+1 \atop j+1} \right\} x^{j+1} \int_{0}^{1} u^{xu} (-u \log u)^{j} \, du. \tag{1}
\end{align*}
To analyze the behavior of
$$ f_{j}(x) = x^{j+1} \int_{0}^{1} u^{xu} (-u \log u)^{j} \, du, $$
we introduce two functions $g_0 : [0, 1/e] \to [0, 1/e]$ and $g_1 : [0, 1/e] \to [1/e, 1]$ ad follows:
The graph of $y = -x \log x$ on $[0, 1]$ can be divided into two parts:

the increasing part and the decreasing part. Thus on each part, we can define the inverse function $g_0$ and $g_1$ as follows:

It is clear from the slope of the graph $y = - x \log x$ that $g_{0}'(0) = 0$ and $g_{1}' (0) = -1$. Also, obviously both $g_{0}'(y)$ and $g_{1}'(y)$ are integrable on $[0, 1/e]$. Keeping these in mind, we make the change of variable as follows:
\begin{align*}
f_{j}(x)
&= x^{j+1} \left[ \int_{0}^{1/e} e^{x u \log u} (-u \log u)^{j} \, du + \int_{1/e}^{1} e^{x u \log u} (-u \log u)^{j} \, du \right] \\
&= x^{j+1} \left[ \int_{0}^{1/e} y^{j} e^{-xy} g_{0}'(y) \, dy + \int_{0}^{1/e} y^{j} e^{-xy} (- g_{1}'(y) ) \, dy \right] \\
&= \int_{0}^{\infty} x^{j+1} y^{j} e^{-xy} \left( g_{0}'(y) \mathrm{1}_{[0, 1/e]}(y) \right) \, dy + \int_{0}^{\infty} x^{j+1} y^{j} e^{-xy} \left( - g_{1}'(y) \mathrm{1}_{[0, 1/e]}(y) \right) \, dy
\end{align*}
Here, observe that the family of mappings
$$ y \mapsto \frac{1}{j!} x^{j+1} y^{j} e^{-xy} $$
serves as an approximation to the identity as $x \to \infty$. This proves that
$$ f_{j}(x) \xrightarrow{x\to\infty} j! (g_{0}'(0) - g_{1}'(0)) = j!. $$
So taking limit $x \to \infty$ to the identity $(1)$, we obtain
\begin{align*}
\lim_{x\to\infty} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^{k-m}} x^{k}
&= \sum_{j=0}^{m} (-1)^{j} \left\{ {m+1 \atop j+1} \right\} j!.
\end{align*}
But it is known that the last sum is equal to 
$$ \sum_{j=0}^{m} (-1)^{j} \left\{ {m+1 \atop j+1} \right\} j! = \delta_{0m} = \begin{cases} 1 & m = 0 \\ 0 & m \geq 1 \end{cases}. $$
Therefore we have
$$ \lim_{x\to\infty} \sum_{k=1}^{\infty} \frac{(-1)^{k}}{k^{k-m}} x^{k} = -\delta_{0m}. $$
The following graphs corresponds to $m = 1, \cdots, 5$ in zigzag order.


Added. Approximation to the identity is a way of justifying the concept of the Dirac delta. Consider a family of functions $\{ K_{t}(x) \}$ (parametrized by $t$ in this example) such that
$$ \int_{\Bbb{R}} K_{t}(x) \,dx = q \quad \text{and} \quad \lim_{t\to\infty} \int_{|x|\geq \eta} K_{t}(x) \, dx = 0 \quad \text{for} \ \eta > 0. $$
That is, we cant interpret $K_{t}(x)$ as a mass density with the unit total mass such that the mass concentrates to the origin as $t \to \infty$. So it is natural to expect that $K_{t}(x)$ converges to the Dirac delta $\delta(x)$ in some sense as $t \to \infty$. The theory of approximation to the identity justifies this naive idea under various assumptions and in various mode of convergence.
In our problem,
$$ K_{x}(y) = x f(xy) \quad \text{where} \quad f(y) = \frac{y^{j}}{j!}e^{-y} \quad \text{on} \ [0, \infty). $$
Thus we expect that as $x \to \infty$,
\begin{align*}
f_{j}(x) 
&= j! \int_{0}^{\infty} \left\{ g_{0}'(y) - g_{1}'(y) \right\} \mathbf{1}_{[0, 1/e]}(y) K_{x}(y) \, dy \\
&= j! \int_{0}^{\infty} \left\{ g_{0}'(y) - g_{1}'(y) \right\} \mathbf{1}_{[0, 1/e]}(y) \delta(y) \, dy \\
&= j! \left\{ g_{0}'(0) - g_{1}'(0) \right\} = j!.
\end{align*}
Indeed, some simple estimation proves this.
A: 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.
A: 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) 

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

