Convergence of $\sum_{n=1}^\infty \frac{1}{n}-\ln(1+ \frac{1}{n})$ I'm having trouble to prove/disprove the Series $$\sum_{n=1}^\infty\left( \frac{1}{n}-\ln(1+ \frac{1}{n})\right)$$ being convergent or divergent.

I tried the ratio test but it seems to be inconclusive. I think I should use here some reference for the comparison test but I couldn't find any matching Series.

Also tried to make a common denominator and received $\sum_{n=1}^\infty \frac{1-n\ln(1+ \frac{1}{n})}{n}$ so I have to find something smaller then $n\ln(1+ \frac{1}{n})$, and I couldn't figure out the matching alternative.

Would appreciate your advice
Note that $\ln (1+x)=x-\frac12x^2+O(x^3)$. Hence $\frac1n-\ln(1+\frac 1n)=\frac1{2n^2}+O(n^{-3})$ and so for $n$ large enough we have $0&lt;\frac1n-\ln(1+\frac 1n)&lt;\frac1{n^2}$.

In case you are "afraid" of the big-O or did not learn about Taylor expansion yet, it is sufficient to show the following:
$$ x-x^2\le \ln(1+x)\le x\qquad\text{for }x\ge0.$$
The right hand side is an immediate consequence of $e^x\ge 1+x$ (the mother of all inequalities for the exponential). The left hand side follows from 
$e^{x^2-x}\ge 1+x^2-x&gt;0$ so that
$$e^{x-x^2}=\frac1{e^{x^2-x}}\le \frac1{1-x+x^2} =\frac{1+x}{1+x^3}\le 1+x$$
Just use the limit comparison test .
Use the series $\sum_{n=1}^\infty(1/n)$.
The limit comes out to be 1 and hence both the series converge or diverge together.