Show that $\sum_{n=1}^{\infty}\frac{\log (1+1/n)}{n}$ converges

Question

I need some help here. I can show that if the Cauchy condensation test holds, then I get two separate series, one which converges by the comparison test, and one that converges by the ratio test. But I don't even know if this is a valid argument, since I'm not sure how to even check that the terms are decreasing. So I don't think this approach works.

I see that the same question has been asked here, but I'm not really satisfied with the answers. Are there simple ways to determine convergence, with something like the comparison test?

Answer 1

By summation by parts

$$ \sum_{n=1}^{N}\frac{\log(1+1/n)}{n}=\frac{\log(N+1)}{N}+\sum_{n=1}^{N-1}\frac{\log(n+1)}{n(n+1)} $$ and by the Cauchy-Schwarz inequality $\log(n+1)\leq \sqrt{n+1}-\frac{1}{\sqrt{n+1}}$, such that the rearranged/decelerated series $\sum_{n\geq 1}\frac{\log(n+1)}{n(n+1)}$ is blatantly absolutely convergent.

By Frullani's theorem we also have the integral representation $$ \sum_{n\geq 1}\frac{\log(n+1)}{n(n+1)}=\int_{0}^{+\infty}\frac{(e^{-x}-1)\log(1-e^{-x})}{x}\,dx=\int_{0}^{1}\frac{(1-x)\log(1-x)}{x\log x}\,dx. $$

Answer 2

$$\log(1+x)\le x$$

implies

$$\sum_{n=1}^\infty\frac{\log\left(1+\dfrac1n\right)}{n}\le \sum_{n=1}^\infty\frac1{n^2}$$

is simple and based on the comparison test.