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Question Number 182940 by mathocean1 last updated on 17/Dec/22

$$Are\sum _{n⩾1}\frac{1}{4n^2-1}and$$$$\sum _{n⩾1}\frac{1}{n(n+1)(n+2)}convergent?$$

Answered by JDamian last updated on 17/Dec/22

$$a_n=\frac{1}{4n^2-1}=\frac{1}{(2n-1)(2n+1)}=$$$$=\frac{1}{2}(\frac{1}{2n-1}-\frac{1}{2n+1})telescopingseries$$$$S=\frac{1}{2}\underset{k=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{2n-1}-\frac{1}{2n+1})=$$$$=\frac{1}{2}(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...)=$$$$=\frac{1}{2}[1-\underset{n\to \mathrm{\infty }}{\lim }\left(\frac{1}{2n+1}\right)]=\frac{1}{2}$$$$$$

Answered by ARUNG_Brandon_MBU last updated on 17/Dec/22

$$S=\sum _{n⩾1}\frac{1}{n(n+1)(n+2)}=\sum _{n⩾1}(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2(n+2)})$$$$=\frac{1}{2}\sum _{n⩾1}\frac{1}{n}-(\sum _{n⩾1}\frac{1}{n}-1)+\frac{1}{2}(\sum _{n⩾1}\frac{1}{n}-1-\frac{1}{2})$$$$=1-\frac{1}{2}(1+\frac{1}{2})=1-\frac{3}{4}=\frac{1}{4}$$

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