I need to prove that:
$$\sum_{n=1}^{\infty} (-1)^{n+1}\log\left(1+\frac{1}{n}\right)$$
is convergent, but not absolutely convergent.
I tried the ratio test:
$$\frac{a_{n+1}}{a_n} = -\frac{\log\left(1+\frac{1}{n+1}\right)}{\log\left(1+\frac{1}{n}\right)} = -\log\left({\frac{1}{n+1}-\frac{1}{n}}\right)$$
I know that the thing inside the $\log$ converges to $1$, so $-\log$ converges to $0$? This is not right, I cannot conclude that this series is divergent.
Also, for the sequence without the $(-1)^{n+1}$ it would give $0$ too.
Best Answer
First, because is an alternating series, we start with the alternating series test: if $(a_n)\to 0$ and $|a_n|$ decreases then $\sum(-1)^{n}a_n$ converges.
We can see that $\log\left(\frac{n+1}{n}\right)\to 0$ because $\left(\frac{n+1}{n}\right)\to 1$ and $\left|\log\left(\frac{n+1}{n}\right)\right|$ decreases because logarithm is a strict increasing function and $\frac{n+1}{n}$ obviously decreases. Then the series, at least, is conditionally convergent.
To test absolute convergence we can see that we can write
$$\sum_{n=1}^{\infty}\log\left(\frac{n+1}{n}\right)=\sum_{n=1}^{\infty}\log(n+1)-\log(n)$$
that is a telescoping sum. And because $\sum_{n=1}^{\infty}a_n=\lim_{N\to\infty}\sum_{n=1}^{N}a_n$ then
$$\sum_{n=1}^{\infty}\log(n+1)-\log(n)=\lim_{N\to\infty}\sum_{n=1}^{N}\log(n+1)-\log(n)=\lim_{N\to\infty}\log(N+1)$$
what is divergent, so the series is conditionally convergent.