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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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