Series Divergence Proof

Question

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!

Answer 1

It's a necessary (but not sufficient) criterion for any series to converge that the sequence converges to $0$. If it's an alternating series in which the absolute values are monotonically decreasing to $0$, it's also sufficient for convergence.

In this case, since your sequence does not converge to $0$, you're done; it diverges. (It doesn't matter if it's alternating or not.)

Answer 2

Suppose you let $c_n=(-1)^n a_n$.

As you've shown, $|c_n|=a_n\not\to0\;,$ so $c_n\not\to0$ also $\;\;$since $c_n\to0\iff|c_n|\to0$.

Therefore you can conclude that $\displaystyle\sum_{n=1}^{\infty}c_n$ diverges by the Divergence Test.

Answer 3

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 $O(1/n^2)$.