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.

$S_n=n\cdot\sum_{j=0}^{\infty}\dfrac {1}{(n+2j)(n+2j+1)}.$ For $n\ge 2$ this is bounded below by $$\frac {n}{2}\sum_{j=0}^{\infty}\int_{n+2j}^{n+2j+2}\frac {1}{x^2}dx=\frac {n}{2}\int_n^{\infty}\frac {1}{x^2}dx=\frac {1}{2}$$ and is bounded above by $$\frac {n}{2}\sum_{j=0}^{\infty}\int_{n+2j-1}^{n+2j+1}\frac {1}{x^2}dx=\frac {n}{2}\int_{n-1}^{\infty}\frac {1}{x^2}dx=\frac {n}{2(n-1)}.$$
