Notice that if $\displaystyle k \ge n+1$, the summand is 0, so the upper limit $\displaystyle n$ is irrelevant. Let it sum to infinity and it is the same sum. Then we can switch the order of the summation signs since now the inner sum does not depend on $\displaystyle n$.
$\displaystyle
\displaystyle = \sum_{n=0}^{\infty}\sum_{k=0}^{n} \binom{n}{k}\frac{x^{n-k}}{(1+x)^{n+1}}\frac{(-1)^k}{(1+k)^s}$
$\displaystyle \displaystyle = \sum_{k=0}^{\infty}\sum_{n=0}^{\infty} \binom{n}{k}\frac{x^{n-k}}{(1+x)^{n+1}}\frac{(-1)^k}{(1+k)^s}$
$\displaystyle \displaystyle = \sum_{k=0}^{\infty} \frac{(-1)^k}{x^k(1+k)^s(1+x)} \sum_{n=0}^{\infty} \binom{n}{k}\left(\frac{x}{1+x}\right)^n$
$\displaystyle \displaystyle = \sum_{k=0}^{\infty} \frac{(-1)^k}{x^k(1+k)^s(1+x)} \frac{\left(\frac{x}{1+x}\right)^k}{\left(1-\frac{x}{1+x}\right)^{k+1}}$
$\displaystyle \displaystyle = \sum_{k=0}^{\infty} \frac{(-1)^k(1+x)x^k}{x^k(1+k)^s(1+x)} \frac{\left(\frac{1}{1+x}\right)^k}{\left(\frac{1} {1+x}\right)^{k}}$
$\displaystyle \displaystyle = \sum_{k=0}^{\infty} \frac{(-1)^k}{(1+k)^s}$
$\displaystyle \displaystyle = \sum_{k=0}^{\infty} \frac{(-1)^{n+1}}{n^s}$
Multiply the last line by the fraction and you get the analytic continuation of $\displaystyle \zeta(s)$ for real part of s > 0.
The sum was done by the formula
$\displaystyle
\displaystyle \sum_{n}\binom{n}{k}y^n = \frac{y^k}{(1-y)^{k+1}}$
here we sum over all values of n. The summand is zero for n < k.