Show that the series $\sum (-1)^n \frac{n}{(n^2 + v_n)}$ converges. Let $\{v_n\}_{n \in\mathbb N}$ be a sequence such that $ 0 \leq v_n \leq n $ for all $n  \in\mathbb N$. Show that the series
$$
\sum (-1)^n  \frac{n}{n^2 + v_n}
$$
converges.
I have this question I'm looking to solve. I've usually solved questions which have terms with 'n' but this one has a sequence in the term and I'm not sure how to begin.
I thought of maybe starting with Absolute Convergence Test and/or Alternating Series Test but I couldn't go too far. Can someone help me figure this one out? 
Thanks!!
 A: It's a little bit of a funky question, but it can be solved using the alternating series test. The bounds imposed on $v_n$ ensure that $\frac{n}{n^2 + v_n}$ converges monotonically to $0$:
\begin{align*}
\frac{n}{n^2 + v_n} - \frac{n+1}{(n+1)^2 + v_{n+1}} &= \frac{n((n+1)^2 + v_{n+1}) - (n+1)(n^2 + v_n)}{((n+1)^2 + v_{n+1})(n^2 + v_n)} \\
&= \frac{n^2 + n + n v_{n+1} - (n+1)v_n}{((n+1)^2 + v_{n+1})(n^2 + v_n)} \\
&\ge \frac{n^2 + n + n v_{n+1} - (n+1)n}{((n+1)^2 + v_{n+1})(n^2 + v_n)} \\
&= \frac{n v_{n+1}}{((n+1)^2 + v_{n+1})(n^2 + v_n)} \\
&\ge 0.
\end{align*}
Thus, $\sum (-1)^n \frac{n}{n^2 + v_n}$ converges (in this case, conditionally).
A: To use Leibniz need to show 
1) $a_n:=\frac {n}{n^2+v_e}$ is decreasing.
$0\lt \frac{n}{n^2+n} \le a_n \le \frac{n}{n^2}=\frac{1}{n}$;
$\frac{1}{n+1} \le a_n \le \frac{1}{n}.$
$a_n-a_{n+1} \ge \frac{1}{n+1} -\frac{1}{1+n} =0$.
2) $0< a_n \le 1/n,$ hence $\lim_{n\rightarrow \infty}a_n=0.$
A: Note that $a_n=\frac{n}{n^2+v_n}$ is decreasing because 
$$
a_n=\frac{n}{n^2+v_n}\geq\frac{n}{n^2+n}=\frac{1}{n+1}=\frac{n+1}{n(n+1)}\geq\frac{n+1}{(n+1)^2+v_{n+1}}=a_{n+1}
$$
Moreover, $\lim\limits_{n\rightarrow\infty}a_n=0$.
