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Question Number 152377 by fotosy2k last updated on 27/Aug/21

Commented by fotosy2k last updated on 28/Aug/21

$$plshelp$$

Commented by Tawa11 last updated on 28/Aug/21

$$\mathrm{good}$$

Answered by Mokmokhi last updated on 28/Aug/21

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{n}-\frac{1}{n+3})$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\left(\frac{1}{n}\right)-\underset{n=1}{\sum ^{\mathrm{\infty }}}\left(\frac{1}{n+3}\right)$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\left(\frac{1}{n}\right)-\underset{n=4}{\sum ^{\mathrm{\infty }}}\left(\frac{1}{n}\right)$$$$=1+\frac{1}{2}+\frac{1}{3}$$$$=\frac{11}{6}$$

Answered by Kamel last updated on 28/Aug/21

$$S=\underset{n=1}{\sum ^{+\mathrm{\infty }}}(\{\frac{1}{n}-\frac{1}{n+1}\}+\{\frac{1}{n+1}-\frac{1}{n+2}\}+\{\frac{1}{n+2}-\frac{1}{n+3}\})$$$$=1+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}$$

Answered by qaz last updated on 28/Aug/21

$${\mathrm{H}}_3=1+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}$$

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