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.
 A: Notice that
$$
x - \ln(1+x) = \int_0^x \frac{t}{1+t}dt
$$ 
hence for all $x \geq 0$,
$$
0 \leq x - \ln(1+x) \leq \int_0^x t dt = \frac{x^2}{2}.
$$
Now, you can use the comparision test.
Edit: if you don't want integrals, then you can prove the same inequality by studying the variations of the functions
$$
f(x) = x - \ln(1+x),\qquad g(x) = x - \ln(1+x) - \frac{x^2}{2}.
$$
A: 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.
A: By MVT, for any positive $x > 0$, there exists a $\xi \in (0,1)$ such that
$$0 \le x - \log(1+x) = \left(1 - \frac{1}{1+\xi x}\right) x = \frac{\xi x^2}{1 + \xi x} \le x^2$$
This implies the series at hand consists of non-negative terms and its sum is bounded from above by
$$\sum_{n=1}^\infty\left(\frac1n - \log\left(1 + \frac1n\right)\right) \le \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$
As a result, the series converges.
