Question: I would like to solve the following series: $$\sum\_{n=3}^\infty \frac{\log(n-1)-\log(n)}{\log(n-1)\log(n)}$$

Here's what I did:
$$\sum\_{n=3}^\infty \frac{\log(n-1)-\log(n)}{\log(n-1)\log(n)}=\sum\_{n=3}^\infty \frac{\log\bigl(\frac{n-1}{n}\bigr)}{\log(n-1)\log(n)}=\sum\_{n=3}^\infty\frac{\log\bigl(1-\frac{1}{n}\bigr)}{\log(n-1)\log(n)}$$

Apparently I can solve this by using the comparison test, but I don't know how this would be implemented in this exercise. Also, if it is possible, how can I find the result of the series?

Answer: 
$$\begin{align}
\sum\_{n=3}^{\infty} \frac{\log(n-1)-\log(n)}{\log(n-1) \log(n)}
&=\sum\_{n=3}^{\infty}\left( \frac{\log(n-1)}{\log(n-1) \log(n)}-\frac{\log(n)}{\log(n-1) \log(n)}\right) \\
&=\sum\_{n=3}^{\infty}\left( \frac{1}{\log(n)}-\frac{1}{\log(n-1)}\right) \\
&=\lim\_{N\to \infty}\sum\_{n=3}^N\left( \frac{1}{\log(n)}-\frac{1}{\log(n-1)}\right) \\
&=\lim\_{N\to \infty}\left(\frac{1}{\log N}-\frac{1}{\log 2}\right) \\
&=-\frac{1}{\log 2}.
\end{align}$$
Hint: It is $$\sum\_{n=3}^{\infty}\frac{1}{\log(n)}-\frac{1}{\log(n-1)}$$