= Solution
The <alternating series test> says that if $a_n\geq0$, $a_{n+1}\leq a_n$, and $a_n\to0$, then $\sum_{n\geq1}(-1)^{n+1}a_n$ converges.
To prove it, let $S_N$ be the partial sums. The even sums satisfy
$$
S_{2m+2}-S_{2m}
=a_{2m+1}-a_{2m+2}\geq0,
$$
so $(S_{2m})$ is increasing. The odd sums satisfy
$$
S_{2m+3}-S_{2m+1}
=-a_{2m+2}+a_{2m+3}\leq0,
$$
so $(S_{2m+1})$ is decreasing. Also $S_{2m}\leq S_{2m+1}$, so both are bounded and converge. Their difference is $a_{2m+1}\to0$, 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. Since
$$
S_4=1-\frac12+\frac13-\frac14=\frac7{12}
$$
and
$$
S_5=S_4+\frac15=\frac{47}{60},
$$
we obtain
$$
\boxed{\frac7{12}\leq S\leq\frac{47}{60}}.
$$
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