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!!

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).

• > The bounds imposed on $v_n$ ensure that $\frac{n}{n^2 + v_n}$ converges monotonically to $0$: Could you please explain what you mean by this? Also, after that, are you trying to show that it $a_n$ $\geq$ $a_{n+1}$ by proving that its $\geq$ 0 and hence is monotonically decreasing so by AST, $\lim\limits_{n\rightarrow\infty}a_n=0$? Commented Apr 26, 2020 at 13:57
• I'm trying to show that $a_n$ is monotonically decreasing. This doesn't prove $a_n \to 0$; that's a separate observation (try squeeze theorem). The fact that it is monotonic, decreasing, and converges to $0$, means that $\sum (-1)^n a_n$ converges by the AST. Commented Apr 26, 2020 at 15:20
• Hey I have a follow up question: how do you prove that $a_n$ $\to$ 0? Commented Apr 29, 2020 at 0:27

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 $$\lim_{n\rightarrow \infty}a_n=0.$$

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, $$\lim\limits_{n\rightarrow\infty}a_n=0$$.