Proving convergence of $\sum_{n=1}^\infty \frac{1}{n} \log(1 + \frac{1}{n})$ I wonder if this is a valid proof that $$\sum_{n=1}^\infty \frac{1}{n} \log\left(1 + \frac{1}{n}\right)$$
converges:
$$\frac{1}{n} \log\left(1 + \frac{1}{n}\right) < \frac{1}{n} \log(1) = \frac{1}{n}\times0 = 0$$
And the sum $\sum_{n=1}^\infty 0 = 0$, therefore from the comparison test the sum from the title ($\sum_{n=1}^\infty \frac{1}{n} \log(1 + \frac{1}{n})$) converges.
Is this valid? I feel like I'm doing something forbidden here by comparing it with 0.
 A: No, this is not valid at all. For every $n$, the summands are positive - they aren't smaller than $0$.
And even if the first inequality was correct, the result wouldn't follow. $\sum_{n = 1}^{\infty} -1 $ is divergent, even though $-1 < 0$.

For an approach that works, recall that $\ln(1 + x) \approx x$ for $x$ small enough.
A: $$\frac{1}{n} \log(1 + \frac{1}{n}) < \frac{1}{n} \log(1)$$
This is not true. Plot a graph to see why.
If you want to prove convergence, perhaps try using this (just an idea - might not work):
$$\log\left(1 + \frac{1}{n}\right)=\log\left(\frac{n+1}{n}\right)$$
$$\log\left(\frac{n+1}{n}\right)=\log\left(n+1\right)-\log\left(n\right)$$
A: hint: use $\log(1+x) \leq x$, with $x = \dfrac{1}{n}$, and the fact that the series $\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^2}$ converges as a $p$-series with $p > 1$
A: 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
$$\bbox[5px,border:2px solid #C0A000]{1 \le \sum_{n=1}^\infty \frac1n\log\left(1+\frac1n\right)\le \frac{\pi^2}{6}}$$
