Example 5. Investigate the convergence of the infinite product

$$
\prod_{k=1}^{\infty}\left(1-\frac{1}{k+1}\right)=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right) \ldots\left(1-\frac{1}{k+1}\right) \ldots
$$
Solution. Here all $u_{k}=-\frac{1}{k+1}$ are negative and the series (14)

$$
\sum_{k=1}^{\infty} u_{k}=-\sum_{k=1}^{\infty} \frac{1}{k+1}=-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots\right)
$$

obviously diverges.

Then, by Theorem 4, the infinite product (15) diverges.

Remark. Calculating the $n$-th partial product in (15), we get

$$
p_{n}=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \ldots \cdot \frac{n-1}{n} \cdot \frac{n}{n+1}=\frac{1}{n+1}
$$

Since $p=\lim _{n \rightarrow \infty} p_{n}=\lim _{n \rightarrow \infty} \frac{1}{n+1}=0$, the infinite product (15) diverges.