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Question Number 11800 by tawa last updated on 01/Apr/17

$$\mathrm{Find}\mathrm{the}\mathrm{sum}$$$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{1}{(\mathrm{n}+1)\sqrt{\mathrm{n}}+\mathrm{n}\sqrt{\mathrm{n}+1}}$$

Answered by FilupS last updated on 01/Apr/17

$$\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})$$$$=(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}})+(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}})+(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}})+...+(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}})+(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})$$$$=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{n+1}}$$$$=1-\frac{1}{\sqrt{n+1}}$$$$\mathrm{as}n\to \mathrm{\infty },\frac{1}{\sqrt{n+1}}\to 0$$$$\therefore =1$$

Commented by FilupS last updated on 01/Apr/17

$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{unable}\mathrm{to}\mathrm{do}\mathrm{the}\mathrm{working}\mathrm{for}\mathrm{line}1.$$$$\mathrm{I}\mathrm{used}\mathrm{WolframAlpha}\mathrm{to}\mathrm{expand}$$

Commented by tawa last updated on 01/Apr/17

$$\mathrm{i}\mathrm{appreciate}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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