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Question Number 144380 by SOMEDAVONG last updated on 25/Jun/21

$$\mathrm{L}=\underset{\mathrm{n}\to +\propto }{\lim }(\frac{1}{1^2+2\left(1\right)}+\frac{1}{2^2+2\mathrm{n}}+...+\frac{1}{{\mathrm{n}}^2+2\mathrm{n}})=?$$

Answered by mathmax by abdo last updated on 25/Jun/21

$${\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{{\mathrm{k}}^2+2\mathrm{k}}\Rightarrow {\mathrm{S}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}(\mathrm{k}+2)}$$$$=\frac{1}{2}\sum _{\mathrm{k}=1}^{\mathrm{n}}(\frac{1}{\mathrm{k}}-\frac{1}{\mathrm{k}+2})=\frac{1}{2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}}-\frac{1}{2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+2}(\mathrm{k}+2=\mathrm{p})$$$$=\frac{1}{2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}}-\frac{1}{2}\sum _{\mathrm{p}=3}^{\mathrm{n}+2}\frac{1}{\mathrm{p}}$$$$=\frac{1}{2}(1+\frac{1}{2}+\sum _{\mathrm{k}=3}^{\mathrm{n}}\frac{1}{\mathrm{k}})-\frac{1}{2}\sum _{\mathrm{k}=3}^{\mathrm{n}}\frac{1}{\mathrm{k}}-\frac{1}{2}(\frac{1}{\mathrm{n}+1}+\frac{1}{\mathrm{n}+2})$$$$=\frac{3}{4}-\frac{1}{2}(\frac{1}{\mathrm{n}+1}+\frac{1}{\mathrm{n}+2})\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{S}}_{\mathrm{n}}=\frac{3}{4}$$

Commented by SOMEDAVONG last updated on 25/Jun/21

$$\mathrm{Thanks}\mathrm{sir}!$$

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