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Question Number 119173 by help last updated on 22/Oct/20

Answered by Olaf last updated on 22/Oct/20

$$\mathrm{S}=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k+2}-\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k+3}$$$$\mathrm{S}=(\frac{1}{3}+\underset{k=2}{\sum ^{\mathrm{\infty }}}\frac{1}{k+2})-\underset{k=2}{\sum ^{\mathrm{\infty }}}\frac{1}{k+2}=\frac{1}{3}$$$$$$

Answered by JDamian last updated on 22/Oct/20

$$Asitisatelescopicseries$$$$\underset{k=1}{\sum ^n}(\frac{1}{k+2}-\frac{1}{k+3})=\frac{1}{3}-\frac{1}{n+3}$$$$\underset{n\to \mathrm{\infty }}{\lim }(\frac{1}{3}-\frac{1}{n+3})=\frac{1}{3}$$

Answered by mathdave last updated on 22/Oct/20

$$solution$$$$\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k+2}-\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k+3}=I=(\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k}-1-\frac{1}{2})-(\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k}-1-\frac{1}{2}-\frac{1}{3})$$$$I=-\frac{3}{2}+\frac{11}{6}=\frac{-9+11}{6}=\frac{2}{6}=\frac{1}{3}$$$$\because \underset{k=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{k+2}-\frac{1}{k+3})=\frac{1}{3}$$

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