# How to prove an alternating series is convergent?

How to prove that the sequence $\sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{\sqrt{n}}$ is convergent? I was trying to find the upper bound and lower bound of the partial sum $s_k$ and use Squeeze Theorem to figure out the limit, but I couldn't find the lower bound for $s_k$. Any suggestions? Thanks!

-
The term is undefined at $n=0$; your sum should start at $n=1$, not at $n=0$. –  Arturo Magidin May 2 '11 at 15:49
–  t.b. May 2 '11 at 15:52
Start by combining terms in pairs: $\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}$ to get something absolutely convergent. –  Ross Millikan May 2 '11 at 15:53

This is an alternating series. Letting $a_n = \frac{1}{\sqrt{n}}$, for $n\geq 1$, then the series is $\sum (-1)^{n-1}a_n$. Notice that the sequence $a_n$ is strictly decreasing, as $a_1\gt a_2\gt a_3\gt\cdots$.

Now, consider the sequences of even and odd terms of the partial sum sequence: \begin{align*} s_1 &= 1\\ s_3 &= 1 - \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}}\right)\\ s_5 &= 1 - \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}}\right) - \left(\frac{1}{\sqrt{4}} - \frac{1}{\sqrt{5}}\right)\\ &\vdots\\ s_{2n+1} &= 1 - \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}}\right) -\cdots -\left(\frac{1}{\sqrt{2n}} - \frac{1}{\sqrt{2n+1}}\right)\\ &\vdots \end{align*} and \begin{align*} s_2 &= \left( 1- \frac{1}{\sqrt{2}}\right)\\ s_4 &= \left(1 - \frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}}\right)\\ s_6 &= \left(1 - \frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right) + \left(\frac{1}{\sqrt{5}} - \frac{1}{\sqrt{6}}\right)\\ &\vdots\\ s_{2n} &= \left(1 - \frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right) + \cdots + \left(\frac{1}{\sqrt{2n-1}} - \frac{1}{\sqrt{2n}}\right)\\ &\vdots \end{align*} We have: $$s_2 \lt s_4 \lt s_6 \lt\cdots \lt s_{2n} \lt \cdots \lt s_{2k+1} \lt \cdots \lt s_3 \lt s_1.$$

The sequence of odd terms of the partial sum sequence is strictly decreasing, and bounded below by each of the even terms, so it converges to some $L$. The sequence of even terms of the partial sum sequence is strictly increasing and bounded above by each of the odd terms, so it converges to some $M$; we have $M\leq L$. But notice that $\lim\limits_{n\to\infty}a_n = 0$, so $$L-M = \lim_{n\to\infty}(s_{2n+1} - s_{2n}) = \lim_{n\to\infty}\left(\sum_{k=1}^{2n+1}(-1)^{k-1}a_k - \sum_{k=1}^{2n}(-1)^{k-1}a_k\right) = \lim_{n\to\infty}(-1)^{2n+1}a_n = 0,$$ so $L=M$. It is now straightforward to verify that the sequence of partial sums $s_n$ converges to this common limit $L$, so the series converges.

(This is just the usual proof that the Alternating Series Test "works).

-

Look up: "alternating series".

-