22. $[\mathbf{1 2}]$ Evaluate the infinite sum
$$
\sum_{n=2}^{\infty} \log _{2}\left(\frac{1-\frac{1}{n}}{1-\frac{1}{n+1}}\right) .
$$
Answer: -1 Using the identity $\log _{2}\left(\frac{a}{b}\right)=\log _{2} a-\log _{2} b$, the sum becomes
$$
\sum_{n=2}^{\infty} \log _{2}\left(\frac{n-1}{n}\right)-\sum_{n=2}^{\infty} \log _{2}\left(\frac{n}{n+1}\right) .
$$

Most of the terms cancel out, except the $\log _{2}\left(\frac{1}{2}\right)$ term from the first sum. Therefore, the answer is $-1$