study the convergence of $\sum_{n=1}^\infty \log \frac{n+1}{n}$ $\sum_{n=1}^\infty \log \frac{n+1}{n}$
$\sum_{n=1}^\infty \log \frac{n+1}{n}$ =  $\displaystyle\lim_{n \to{+}\infty}(\log2 - \log1)+(\log3-\log2)+...+(\log(n+1)-\log n)$=$\displaystyle\lim_{n \to{+}\infty}\log(n+1)\to \infty$.
So, the series diverges.
Is my procedure correct?
 A: In the comments, @Marvis shows how to clean up your proof. 
Here's another approach. 
Let $f_n = \log \frac{n+1}{n}$. 
Examine the ratio of successive terms for large $n$, 
$$\frac{f_{n+1}}{f_{n}} = 1 - \frac{1}{n} + O\left(\frac{1}{n^2}\right).$$ 
Therefore, the series diverges by Gauss's test. 
A: Your procedure is fine.  I think it easiest to remove the logarithm immediately:
$e^{a_k} = \prod_{n=1}^k{\frac{n+1}{n}} = k+1$
So, $a_k = \log{(k+1)} \rightarrow \infty$.
A: Your procedure is correct. I suggest another way:
$\log \frac{n+1}{n}$ is positive, and: $\log \left(1+\frac{1}{n}\right)\sim \frac{1}{n}$ for $n \rightarrow\infty$. So, we have:
$\sum \frac{1}{n}$  that diverges.
A: Your procedure is correct. If you want to write out things more clearly I suggest that you write down the $n$th partial sums
$$\begin{align}
s_n &= \sum_{i=1}^{n} \log\left(\frac{i+1}{i}\right)  \\
&= \sum_{i=1}^{n} \log(i+1) - \log(i) \\
&= [\log(2) - \log(1)] + \dots [\log(n+1) - \log(n)]  \\
&= \log(n+1).
\end{align}
$$
Hence 
$$
\lim_{n \to \infty} s_n = \lim_{n\to \infty} \log(n+1) = \infty.$$
So then you say that since the limit does not exist, the series is divergent by definition.
Note: The notation is important. It is not correct to write $\lim_{n\to \infty} \log(n+1) \to \infty$, we write $\lim_{n\to \infty} \log(n+1) = \infty.$
