Update: edited text and formatting.
We can write the sum $e_{n,k}$ in the form (see section 4 of this article by Alf van der Poorten)
$$e_{n,k}=\sum_{m=1}^{k}\frac{(-1)^{m-1}}{2m^{3}\dbinom{n}{m}\dbinom{n+m}{m}},
\quad (1\leq k\leq n).$$
Let $u_{n,m}=m^{3}\binom{n}{m}\binom{n+m}{m}$. If $1=m\leq n$, then $u_{n,m}=n\left( n+1\right)$. To show that $u_{n,m}>n(n+1)$ for $1<m\leq n$ we consider the following two cases:
if $1<m=n$, then $u_{n,m}=n^{3}\binom{2n}{n}>n(n+1)$;
if $1<m\leq n-1$, then $m^{3}\binom{n}{m}\geq m^{3}\binom{n}{1}=m^{3}n>n$ and $\binom{n+m}{m}\geq \binom{n+m}{1}=n+m>n+1$. Hence $u_{n,m}>n(n+1)$.
Hence, for $1<k\leq n$, we get:
$$\begin{eqnarray*}
\left\vert e_{n,k}\right\vert &=&\left\vert \sum_{m=1}^{k}\frac{(-1)^{m-1}}{%
2m^{3}\binom{n}{m}\binom{n+m}{m}}\right\vert \leq \sum_{m=1}^{k}\left\vert
\frac{(-1)^{m-1}}{2m^{3}\binom{n}{m}\binom{n+m}{m}}\right\vert\leq \sum_{m=1}^{n}\frac{1}{2m^{3}\binom{n}{m}\binom{n+m}{m}}\\&<&\sum_{m=1}^{n}\frac{1}{2n(n+1)}=\frac{n}{2n(n+1)}<\frac{1}{n}.
\end{eqnarray*}$$
For $k=1$, we get $\left\vert e_{n,1}\right\vert =\frac{1}{2n\left( n+1\right) }%
\leq \frac{1}{2(n+1)}<\frac{1}{n}$. Thus for every integer $1\leq k\leq n$, we proved that $\left\vert e_{n,k}\right\vert <\frac{1}{n}$, which implies that $e_{n,k}$ converges uniformly in $k$ to $0$.