= Solution
Let $S_N=\sum_{n=1}^N(-1)^{n+1}/n$. Integrate the finite <geometric series> identity $\sum_{k=0}^{N-1}(-t)^k=[1-(-t)^N]/(1+t)$ over $0\leq t\leq1$:
$$
S_N=\int_0^1\frac{1-(-t)^N}{1+t}\,dt=\ln2-(-1)^N\int_0^1\frac{t^N}{1+t}\,dt.
$$
The remainder has <absolute value> at most $\int_0^1t^Ndt=1/(N+1)\to0$. Hence the <partial sums> converge, with
$$
\boxed{\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}n=\ln2.}
$$
This proves both convergence and the value of the <alternating harmonic series> without substituting an endpoint into an unproved infinite <power series>. Its convergence is conditional because the corresponding <harmonic series> of <absolute values> diverges.
Back to article page
