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.

Without generating function, the problem is relatively simple.
Consider the partial sum
$$S_p=\sum_{k = 0}^p (-1)^k \frac{n}{k+n}=\sum_{k = 0}^p  \frac{n}{ 2k+n}-\sum_{k = 0}^p  \frac{n}{ 2k+n+1} $$ Using generalized harmonic numbers and/or the polygamma function (I mix them here on purpose)
$$S_p=\frac{n}{2}  \left(\psi \left(\frac{n}{2}+p+1\right)-\psi
   \left(\frac{n}{2}\right)\right)-\frac{n}{2}  \left(H_{\frac{n+1}{2}+p}-H_{\frac{n-1}{2}}\right)$$
Now, using the asymptotics,
$$S_p=\frac{n}{2}  \left(\psi \left(\frac{n+1}{2}\right)-\psi
   \left(\frac{n}{2}\right)\right)-\frac{n}{4 p}+O\left(\frac{1}{p^2}\right)$$ Repeat with
$$\frac{n}{2}  \left(\psi \left(\frac{n+1}{2}\right)-\psi
   \left(\frac{n}{2}\right)\right)=\frac{1}{2}+\frac{1}{4 n}-\frac{1}{8 n^3}+O\left(\frac{1}{n^5}\right)$$
