As others have written, the inequality in the OP is not correct. In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{\frac{x}{x+1}\le \log(1+x)\le x} \tag 1$$
Now, letting $x=1/n$ in $(1)$, we obtain
$$\bbox[5px,border:2px solid #C0A000]{\frac{1}{n(n+1)}\le \frac1n\log\left(1+\frac1n\right)\le \frac1{n^2}} \tag 2$$
So, by the comparison test (or the integral test), the series $\sum_{n=1}^\infty \frac1n\log\left(1+\frac1n\right)$ converges.
Note that we can even obtain from $(2)$ crude bounds for the value of the series of interest. The left-hand side of $(2)$ can be written $\frac{1}{n(n+1)}=\frac1{n}-\frac{1}{n+1}$, which is a telescoping sequence. Therefore, the sum of the left-hand side is $\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=1$.
The sum of the right-hand side was the Basel Problem, first solved by Euler in 1734. The sum is $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$
Putting it all together we have
$$\bbox[5px,border:2px solid #C0A000]{1 \le \sum_{n=1}^\infty \frac1n\log\left(1+\frac1n\right)\le \frac{\pi^2}{6}}$$