Convergence of $\lim_{N \to \infty} \sum_{n=1}^N \ln\left (1 + \frac{1}{n} \right )$ $$\lim_{N \to \infty} \sum_{n=1}^N \ln \left (1 + \frac{1}{n} \right )$$
I am trying to show that is expression converges. This is in the form of a partial sum, so this expression converges if the following converges
$$ \sum_{n=1}^\infty \ln \left ( 1 + \frac{1}{n} \right )$$
I feel like this must be a known result, I am just blanking on how to do it. My usual convergence criteria don't seem to be uselful, besides perhaps this one, about integrability that I'm still not so sure how to use.
For a positive, continuous and decreasing function ( which is what we have ), $\forall x \ge N$, and for $f(n)=a_n$, $\forall n \ge N$, then
$$\sum_{n=1}^\infty a_n\ \text{converges} \iff \lim_{n \to \infty} \int_N^n f(x)\,dx \ \text{exists}$$
In my course we are just getting into integrals so I'm not quite sure how to show that the limit of this integral exists. Is this the right direction to go or should it be simpler to show the convergence of this series?
Update
This was a sub-question to a larger question, which was whether or not the following converges to zero or not. Again, we have a zero times infinity situation. I was hoping just the main term converged.
$$\lim_{N \to \infty} \frac{1}{N} \sum_{n=1}^N \ln \left (1 + \frac{1}{n} \right )$$
 A: But, the aforementioned sum diverges.
For $N$ large enough $\ln(1+\frac{1}{N}) \ge \frac{1}{4N}$ [work out the Taylor expansion to see for yourself; $\ln(1+a) \ge a-a^2/2$], and $\sum_{N} \frac{1}{4N}$ diverges.
A: To prove divergence, note that
$$\ln(1 + 1/n) = \int_1^{1+1/n} \frac{dt}{t}  \geqslant \frac{1/n}{1+1/n} = \frac{1}{n+1}$$
Regarding the update:
If $a_n \to 0$ as $n \to \infty$ then it follows that $\frac{1}{N}\sum_{n=1}^N a_n \to 0 $ as $N \to \infty$.  In this case $a_n = \ln(1+1/n) \to 0 $.
For proof note that for any $\epsilon> 0$ there exists $M$ such that $|a_n| < \epsilon$ for $n > M$ and
$$\left|\frac{1}{N} \sum_{n=1}^N a_n \right| \leqslant \left|\frac{1}{N} \sum_{n=1}^M a_n \right|+ \left|\frac{1}{N} \sum_{n=M+1}^N a_n \right| \leqslant \frac{1}{N} \left|\sum_{n=1}^M a_n \right|+ \frac{N-M}{N}\epsilon$$
Thus,
$$0 \leqslant \liminf_{N \to \infty}\left|\frac{1}{N} \sum_{n=1}^N a_n \right| \leqslant \limsup_{N \to \infty}\left|\frac{1}{N} \sum_{n=1}^N a_n \right| < \epsilon,$$
which implies since $\epsilon$ can be arbitrarily close to $0$,
$$\lim_{N \to \infty}\frac{1}{N} \sum_{n=1}^N a_n = 0$$
A: hint :  
$\ln(1+1/n)= \ln((n+1)/n)= \ln(n+1)-\ln(n)$ .. the sum telescopes
A: $$\ln\frac21+\ln\frac32+\ln\frac43+\cdots\ln\frac{N+1}N=\ln\frac{2\cdot3\cdot4\cdots(N+1)}{1\cdot2\cdot3\cdots N}=\ln(N+1)$$
leaves no doubt about convergence.
