 Solution

The alternating series test says that if $a_n\ge 0$, $a_{n+1}\le a_n$, and $a_n\to 0$, then ${\sum }_{n\ge 1}(-1{)}^{n+1}{a}_n$ converges.

To prove it, let $S_N$ be the partial sums. The even sums satisfyso $(S_{2m})$ is increasing. The odd sums satisfyso $(S_{2m+1})$ is decreasing. Also $S_{2m}\le S_{2m+1}$, so both are bounded and converge. Their difference is $a_{2m+1}\to 0$, hence their limits agree and the whole sequence $(S_N)$ converges.

Taking $a_n=1/n$ proves convergence of the alternating harmonic series. Its even partial sums lie below its limit $S$, while its odd partial sums lie above it. Sinceandwe obtain