Show that $\lim_{n \to \infty} \sum_{k = 0}^\infty (-1)^k \frac{n}{n + k} = \frac{1}{2}$. 
Show that
$$ \lim_{n \to \infty} \sum_{k = 0}^\infty (-1)^k \frac{n}{n + k} = \frac{1}{2}. $$

Progress:
I rewrote the inside as
$$ S_n =  \sum_{k = 0}^\infty (-1)^k \frac{n}{n + k} = \sum_{k = 0}^\infty (-1)^k \left( 1 - \frac{k}{n + k} \right),$$
at which point I sought to find the (ordinary) generating function
$$A(x) = \sum_{k = 0}^\infty (-1)^k \left( 1 - \frac{k}{n + k} \right) x^k = \frac{1}{1 + x} - \sum_{k = 0}^\infty (-1)^k \left( \frac{k}{n + k} \right) x^k  , $$
because then $S_n = A(1)$. Unfortunately, I have been trying to evaluate the last sum using the Taylor Series expansion of $\ln \left( 1 + x \right)$ at $x = 0$, but I am stuck.
While is true that
$$\ln \left( 1 + x \right) = \sum_{k = 1}^\infty (-1)^{k + 1} \left( \frac{1}{k} \right) x^k $$
(for $x \in (-1, 1)$ ; note this is irrelevant if we regard all g.f. as "formal"), I can't figure out how to transform this into $\sum_{k = 1}^\infty (-1)^{k + 1} \left( \frac{1}{n + k} \right) x^k$. The motivation is that once I can find that sum in explicit form, I can differentiate it w.r.t $x$ to find the desired sum. I guess my question is also: "How to shift backwards the coefficients of an ordinary generating function?", though I am eager to see other (possibly more elegant) solutions as well.
Note:
One attempt I tried at shifting the coefficients was dividing the coefficients by some power  $x^k$, but this introduces negative coefficients, whereas I need those terms with negative coefficients to disappear.

If you are willing to accept a solution that doesn't use generating functions, then here are such ones:

1st Solution. Fusing each consecutive even and odd terms into a single term,
\begin{align*}
S_n := \sum_{k=0}^{\infty} (-1)^k \frac{n}{n + k}
&= \sum_{k=0}^{\infty} \left[ \frac{n}{n + 2k} - \frac{n}{n + 2k+1} \right] \\
&= \sum_{k=0}^{\infty} \frac{n}{(n + 2k)(n + 2k+1)} \\
&= \sum_{k=0}^{\infty} \frac{1}{(1 + \frac{2k}{n})(1 + \frac{2k+1}{n})} \cdot \frac{1}{n}.
\end{align*}
The last sum looks like a Riemann sum of $\frac{1}{(1+2x)^2}$ over $[0, \infty)$, and so, it is not unreasonable to expect that $S_n$ will converge to  $\int_{0}^{\infty} \frac{1}{(1+2x)^2} \, \mathrm{d}x = \frac{1}{2}$.
Indeed, noting that $x \mapsto \frac{1}{1+2x}$ is decreasing for $x > -\frac{1}{2}$, we find that
$$
\int_{\frac{k+1}{n}}^{\frac{k+2}{n}} \frac{1}{(1+2x)^2} \, \mathrm{d}x
\leq \frac{1}{(1 + \frac{2k}{n})(1 + \frac{2k+1}{n})} \cdot \frac{1}{n}
\leq \int_{\frac{k-1}{n}}^{\frac{k}{n}} \frac{1}{(1+2x)^2} \, \mathrm{d}x
$$
for $n \geq 3$ and $k \geq 0$. (This condition is just to ensure that $\frac{k-1}{n} > -\frac{1}{2}$.) Altogether,
$$ 
\int_{\frac{1}{n}}^{\infty} \frac{1}{(1+2x)^2} \, \mathrm{d}x
\leq S_n
\leq \int_{-\frac{1}{n}}^{\infty} \frac{1}{(1+2x)^2} \, \mathrm{d}x $$
Letting $n \to \infty$, we conclude:
$$ \lim_{n\to\infty} S_n = \int_{0}^{\infty} \frac{1}{(1+2x)^2} \, \mathrm{d}x = \frac{1}{2}. $$

2nd Solution. We have
\begin{align*}
S_n
&= \lim_{K \to \infty} \sum_{k=0}^{K-1} (-1)^k n \int_{0}^{1} x^{n+k-1} \, \mathrm{d}x \\
&= \lim_{K \to \infty} \int_{0}^{1} \sum_{k=0}^{K-1} n (-1)^k x^{n+k-1} \, \mathrm{d}x \\
&= \lim_{K \to \infty} \int_{0}^{1} n x^{n-1} \frac{1 - (-x)^K}{1 + x} \, \mathrm{d}x \\
&= \int_{0}^{1} \frac{n x^{n-1}}{1 + x} \, \mathrm{d}x \tag{DCT} \\
&= \int_{0}^{1} \frac{1}{1 + y^{1/n}} \, \mathrm{d}y \tag{$y = x^n$} \\
&\to \int_{0}^{1} \frac{1}{1 + y^{0}} \, \mathrm{d}y \quad \text{as } n \to \infty \tag{DCT} \\
&= \frac{1}{2}.
\end{align*}

3rd Solution. Let $f(x) = \frac{1}{1+x}$. Then
\begin{align*}
S_n
= \sum_{k=0}^{\infty} (-1)^k f\bigl(\tfrac{k}{n}\bigr)
&= \sum_{k=0}^{\infty} \left[ f\bigl(\tfrac{2k}{n}\bigr) - f\bigl(\tfrac{2k+1}{n}\bigr) \right] \\
&= - \sum_{k=0}^{\infty} \int_{0}^{\infty} f'(x) \mathbf{1}_{[\frac{2k}{n}, \frac{2k+1}{n}]}(x) \, \mathrm{d}x \\
&= -\int_{0}^{\infty} f'(x) k_n(x) \, \mathrm{d}x, \tag{Tonelli}
\end{align*}
where $k_n(x) = \sum_{k=0}^{\infty} \mathbf{1}_{[\frac{2k}{n}, \frac{2k+1}{n}]}(x)$. Now let $K_n(x) = \int_{0}^{x} k_n(t) \, \mathrm{d}t$. Clearly, $K_n(x) \leq x$ for all $x \geq 0$. Moreover, as $n \to \infty$ it is not hard to see that $K_n(x) \to \frac{1}{2}x$. Then by integration by parts,
\begin{align*}
S_n
&= \int_{0}^{\infty} f''(x) K_n(x) \, \mathrm{d}x \\
&\to \int_{0}^{\infty} f''(x) \cdot \frac{1}{2} x \, \mathrm{d}x \tag{DCT} \\
&= \left[ f'(x) \cdot \frac{1}{2} x \right]_{0}^{\infty} - \left[ \frac{1}{2}f(x) \right]_{0}^{\infty} \\
&= \frac{1}{2}f(0)
= \frac{1}{2}.
\end{align*}
This result generalizes fairly easily to a more general class of functions $f$.
