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.

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
$$
f(x) = x - \ln(1+x),\qquad g(x) = x - \ln(1+x) - \frac{x^2}{2}.
$$
