Question

# We need to find the sum of the given series. sum_{n=4}^infty(frac{1}{n+1}-frac{1}{n+2})

Series
We need to find the sum of the given series.
$$\sum_{n=4}^\infty(\frac{1}{n+1}-\frac{1}{n+2})$$

2021-02-01
We can find the sum by expanding the given series.
$$\sum_{n=4}^\infty(\frac{1}{n+1}-\frac{1}{n+2})=\lim_{k\to\infty}\sum_{n=4}^k(\frac{1}{n+1}-\frac{1}{n+2})$$
$$=\lim_{k\to\infty}[(\frac{1}{4+1}-\frac{1}{4+2})+(\frac{1}{5+1}-\frac{1}{5+2})+(\frac{1}{6+1}-\frac{1}{6+2})+...+(\frac{1}{k+1}-\frac{1}{k+2})]$$
$$=\lim_{k\to\infty}[(\frac{1}{5}-\frac{1}{6})+(\frac{1}{6}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{8})+...+(\frac{1}{k+1}-\frac{1}{k+2})]$$
$$=\lim_{k\to\infty}[(\frac{1}{5}+(-\frac{1}{6}+\frac{1}{6})+(-\frac{1}{7}+\frac{1}{7})+(-\frac{1}{8}+\frac{1}{8})+...+(-\frac{1}{k+1}+\frac{1}{k+1})-\frac{1}{k+2}]$$
$$=[\frac{1}{5}+(-\frac{1}{6}+\frac{1}{6})+(-\frac17+\frac17)+(-\frac18+\frac18)+...+(-\frac{1}{k+1}+\frac{1}{k+1})-\frac{1}{k+2}]$$
$$\lim_{k\to\infty}[\frac{1}{5}+0+0+0+...+0-\frac{1}{k-2}]$$
$$\lim_{k\to\infty}[\frac{1}{5}-\frac{1}{k+2}]$$
$$=\frac15-\frac{1}{\infty+2}=\frac15-0=\frac15$$
$$\sum_{n=4}^\infty(\frac{1}{n+1}-\frac{1}{n+2})=\frac15$$