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

Find the sum Σ_(n=1) ^∞ (1/((n + 1)(√n) + n(√(n + 1))))

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left(\mathrm{n}\:+\:\mathrm{1}\right)\sqrt{\mathrm{n}}\:\:+\:\:\mathrm{n}\sqrt{\mathrm{n}\:+\:\mathrm{1}}} \\ $$

Answered by FilupS last updated on 01/Apr/17

(1/((n+1)(√n)+n(√(n+1))))=(1/(√n))−(1/(√(n+1))) ⇒Σ_(n=1) ^∞ ((1/(√n))−(1/(√(n+1)))) =((1/(√1))−(1/(√2)))+((1/(√2))−(1/(√3)))+((1/(√3))−(1/(√4)))+...+((1/(√(n−1)))−(1/(√n)))+((1/(√n))−(1/(√(n+1)))) =(1/(√1))−(1/(√(n+1))) =1−(1/(√(n+1))) as n→∞, (1/(√(n+1)))→0 ∴=1

$$\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\sqrt{{n}}+{n}\sqrt{{n}+\mathrm{1}}}=\frac{\mathrm{1}}{\sqrt{{n}}}−\frac{\mathrm{1}}{\sqrt{{n}+\mathrm{1}}} \\ $$$$\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\sqrt{{n}}}−\frac{\mathrm{1}}{\sqrt{{n}+\mathrm{1}}}\right) \\ $$$$=\left(\frac{\mathrm{1}}{\sqrt{\mathrm{1}}}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)+\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)+\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}−\frac{\mathrm{1}}{\sqrt{\mathrm{4}}}\right)+...+\left(\frac{\mathrm{1}}{\sqrt{{n}−\mathrm{1}}}−\frac{\mathrm{1}}{\sqrt{{n}}}\right)+\left(\frac{\mathrm{1}}{\sqrt{{n}}}−\frac{\mathrm{1}}{\sqrt{{n}+\mathrm{1}}}\right) \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{1}}}−\frac{\mathrm{1}}{\sqrt{{n}+\mathrm{1}}} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\sqrt{{n}+\mathrm{1}}} \\ $$$$\mathrm{as}\:\:{n}\rightarrow\infty,\:\:\:\frac{\mathrm{1}}{\sqrt{{n}+\mathrm{1}}}\rightarrow\mathrm{0} \\ $$$$\therefore=\mathrm{1} \\ $$

Commented by FilupS last updated on 01/Apr/17

I′m unable to do the working for line 1. I used WolframAlpha to expand

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

Commented by tawa last updated on 01/Apr/17

i appreciate sir. God bless you sir.

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

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