Question

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

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asked 2021-01-31
We need to find the sum of the given series.
\(\sum_{n=4}^\infty(\frac{1}{n+1}-\frac{1}{n+2})\)

Answers (1)

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