Convergence of $\sum^\infty_{n=1} \left(\frac 1 n-\ln(1+\frac 1 n)\right)$ 
Does the series: $\displaystyle\sum^\infty_{n=1} \left(\frac 1 n-\ln(1+\frac 1 n)\right)$ converge?

We know that $\ln (1+x) <x$ for $x>0$ so this series behaves like $\frac 1 n - \frac 1 n$ which is $0$, but that's just intuition. 
I think the only test that would work is the limit comparison test, but I can't find an expression that would work. Any hints?
Note: no integral test nor integrals, nor Taylor, nor Zeta.

Since 
$$\lim_{n\to \infty} n^2\left(\frac{1}{n} - \log\left(1 + \frac{1}{n}\right)\right) = \lim_{x\to 0} \frac{x - \log(1 + x)}{x^2} = \lim_{x\to 0} \dfrac{\frac{x}{1 + x}}{2x} = \lim_{x\to 0} \frac{1}{2(1 + x)} = \frac{1}{2}$$
and $\sum_{n = 1}^\infty \frac{1}{n^2}$ converges, by limit comparison $\sum_{n = 1}^\infty \left(\frac{1}{n} - \log\left(1 + \frac{1}{n}\right)\right)$ converges.
