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Question Number 98087 by Algoritm last updated on 11/Jun/20

Answered by mr W last updated on 11/Jun/20

$$a_k=\frac{1}{{(k+2)}^2+k}=\frac{1}{(k+1)(k+4)}=\frac{1}{3}(\frac{1}{k+1}-\frac{1}{k+4})$$$$\underset{k=1}{\sum ^n}{a}_k=\frac{1}{3}(\underset{k=1}{\sum ^n}\frac{1}{k+1}-\underset{k=1}{\sum ^n}\frac{1}{k+4})$$$$=\frac{1}{3}(\underset{k=2}{\sum ^{n+1}}\frac{1}{k}-\underset{k=5}{\sum ^{n+4}}\frac{1}{k})$$$$=\frac{1}{3}(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{n+2}-\frac{1}{n+3}-\frac{1}{n+4}+\underset{k=5}{\sum ^{n+4}}\frac{1}{k}-\underset{k=5}{\sum ^{n+4}}\frac{1}{k})$$$$=\frac{1}{3}(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{n+2}-\frac{1}{n+3}-\frac{1}{n+4})$$$$\underset{n\to \mathrm{\infty }}{\lim }\underset{k=1}{\sum ^n}{a}_k=\frac{1}{3}(\frac{1}{2}+\frac{1}{3}+\frac{1}{4})=\frac{13}{36}$$

Commented by Algoritm last updated on 11/Jun/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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