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Question Number 122458 by benjo_mathlover last updated on 17/Nov/20

lim_(n→∞) (1/( (√n) )) ((1/( (√1)+(√3)))+(1/( (√3)+(√5)))+...+(1/( (√(2n−1))+(√(2n+1)))) )=?

$$\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{{n}}\:}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}}+\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}}+...+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{n}−\mathrm{1}}+\sqrt{\mathrm{2}{n}+\mathrm{1}}}\:\right)=? \\ $$

Answered by liberty last updated on 17/Nov/20

lim_(n→∞) (1/( (√n))) (Σ_(k=1) ^n (1/( (√(2k−1))+(√(2k+1))))) = lim_(n→∞) (1/( (√n))) (Σ_(k=1) ^n (((√(2k−1))−(√(2k+1)))/(−2)) )= lim_(n→∞) (1/(2(√n))) Σ_(k=1) ^n ((√(2k+1)) −(√(2k−1)) ) = lim_(n→∞) (((√(2n+1))−1)/(2(√n))) = lim_(n→∞) (((√n) ((√(2+(1/n)))−(1/( (√n)))))/(2(√n)))= ((√2)/2).

$$\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}}}\:\left(\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2k}−\mathrm{1}}+\sqrt{\mathrm{2k}+\mathrm{1}}}\right)\:= \\ $$$$\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}}}\:\left(\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\:\frac{\sqrt{\mathrm{2k}−\mathrm{1}}−\sqrt{\mathrm{2k}+\mathrm{1}}}{−\mathrm{2}}\:\right)= \\ $$$$\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{n}}}\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\sqrt{\mathrm{2k}+\mathrm{1}}\:−\sqrt{\mathrm{2k}−\mathrm{1}}\:\right)\:= \\ $$$$\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2n}+\mathrm{1}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{n}}}\:=\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{n}}\:\left(\sqrt{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{n}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}}}\right)}{\mathrm{2}\sqrt{\mathrm{n}}}=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}. \\ $$

Answered by Dwaipayan Shikari last updated on 17/Nov/20

(1/( 2(√n)))((√3)−(√1)+(√5)−(√3)+....+(√(2n+1))−(√(2n−1))) (1/(2(√n)))((√(2n+1))−1)=(1/2)((√(2+(1/n)))−(1/(2(√n))))=(1/( (√2)))

$$\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{{n}}}\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{1}}+\sqrt{\mathrm{5}}−\sqrt{\mathrm{3}}+....+\sqrt{\mathrm{2}{n}+\mathrm{1}}−\sqrt{\mathrm{2}{n}−\mathrm{1}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{{n}}}\left(\sqrt{\mathrm{2}{n}+\mathrm{1}}−\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{2}+\frac{\mathrm{1}}{{n}}}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{{n}}}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$

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