Series Divergence Proof I'm asked to decide whether the following series converges or diverges: 
$$1-\frac{3}{4}+\frac{4}{6}-\frac{5}{8}+\frac{6}{10}-\frac{7}{12}+\cdots$$
So I first looked at $(a_n)=\frac{n+1}{2n}$. Then by separating this into $a_n=\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{n}$, we can say that since we know $\frac{1}{n}\rightarrow0$, then $a_n \rightarrow \frac{1}{2}+\frac{1}{2}\cdot0=\frac{1}{2}$. Then since $(a_n)$ does not converge to $0$, we know that $\sum a_n$ diverges by the divergence test. 
Now I'm stuck. How do I use this to account for the alternating sequence $(-1)^n a_n$? Is there a theorem that would help?
Thanks!

I would like to
abuse terminology somewhat
by saying that
this series
"multi-converges",
because there are
a finite number of values
$v_k$
and a finite set of disjoint sequences
$u_k(n)$ which together make up
the non-negative integers
 such that
$\sum_{i=0}^{u_k(n)} a_i
\to v_k
$.
In this case,
$k=2$
and the sequences
are the even and odd integers.
It is not hard to show that
$\sum_{i=0}^{2n} a_i
$
and
$\sum_{i=0}^{2n+1} a_i
$
each converge
as $n \to \infty$,
but to separate values.
Since
$a_n 
= (-1)^n\frac{n+2}{2n+2}
$,
$\begin{array}\\
a_{2n}+a_{2n+1}
&=\frac{2n+2}{4n+2}-\frac{2n+3}{4n+4}\\
&=\frac12(\frac{2n+2}{2n+1}-\frac{2n+3}{2n+2})\\
&=\frac12(\frac{(2n+2)^2-(2n+1)(2n+3)}{(2n+1)(2n+2)})\\
&=\frac12(\frac{4n^2+8n+4-(4n^2+8n+3)}{(2n+1)(2n+2)})\\
&=\frac12(\frac{1}{(2n+1)(2n+2)})\\
\end{array}
$
and
$\begin{array}\\
a_{2n-1}+a_{2n}
&=-\frac{2n+1}{4n}+\frac{2n+2}{4n+2}\\
&=\frac12(-\frac{2n+1}{2n}+\frac{2n+2}{2n+1})\\
&=\frac12(\frac{-(2n+1)^2+(2n)(2n+2)}{2n(2n+1)})\\
&=\frac12(\frac{-(4n^2+4n+1)+(4n^2+4n}{2n(2n+1)})\\
&=\frac12(\frac{-1}{2n(2n+1)})\\
\end{array}
$
and the sum of each of these
converges
since they are both
$O(1/n^2)$.
