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Question Number 145025 by mathdanisur last updated on 01/Jul/21

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

Answered by phally last updated on 01/Jul/21

$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{(\mathrm{n}+1)\sqrt{\mathrm{n}}-\mathrm{n}\sqrt{\mathrm{n}+1}}{{\left[(\mathrm{n}+1)\sqrt{\mathrm{n}}\right]}^2-{\left(\mathrm{n}\sqrt{\mathrm{n}+1}\right)}^2}$$$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{(\mathrm{n}+1)\sqrt{\mathrm{n}}-\mathrm{n}\sqrt{\mathrm{n}+1}}{\sqrt{{\mathrm{n}}^2}({\mathrm{n}}^2+2\mathrm{n}+1)-{\mathrm{n}}^2(\mathrm{n}+1)}$$$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{(\mathrm{n}+1)\sqrt{\mathrm{n}}-\mathrm{n}\sqrt{\mathrm{n}+1}}{{\mathrm{n}}^3+{2\mathrm{n}}^2+\mathrm{n}-{\mathrm{n}}^3-{\mathrm{n}}^2}$$$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{(\mathrm{n}+1)\sqrt{\mathrm{n}}-\mathrm{n}\sqrt{\mathrm{n}+1}}{\mathrm{n}(\mathrm{n}+1)}$$$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{\sqrt{\mathrm{n}}}-\frac{1}{\sqrt{\mathrm{n}+1}})$$$$=1-\frac{1}{\sqrt{\mathrm{n}+1}}$$

Commented by mathdanisur last updated on 01/Jul/21

$$thankyouSir,answer?$$

Commented by JDamian last updated on 01/Jul/21

you only have to get now the limit when n --> inf

Answered by mathmax by abdo last updated on 01/Jul/21

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

Commented by mathdanisur last updated on 01/Jul/21

$$thanksSir,answer1-\frac{1}{\sqrt{n+1}}?$$

Commented by mathmax by abdo last updated on 01/Jul/21

$$\mathrm{no}\mathrm{answer}\mathrm{is}1$$

Commented by mathdanisur last updated on 01/Jul/21

$$thanksSir$$

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