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Question Number 95401 by john santu last updated on 25/May/20

lim_(n→∞) n^(3/2) {(√(n+1))+(√(n−1))−2(√n) }

$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}n}^{\frac{\mathrm{3}}{\mathrm{2}}} \left\{\sqrt{\mathrm{n}+\mathrm{1}}+\sqrt{\mathrm{n}−\mathrm{1}}−\mathrm{2}\sqrt{\mathrm{n}}\:\right\}\: \\ $$

Answered by john santu last updated on 25/May/20

Commented by bobhans last updated on 25/May/20

great

$$\mathrm{great} \\ $$

Answered by abdomathmax last updated on 25/May/20

let U_n =n^(3/2) {(√(n+1))+(√(n−1))−2(√n)} ⇒ U_n =n^(3/2) {(√n)(√(1+(1/n)))+(√n)(√(1−(1/n)))−2(√n)} but (√(1+(1/n)))=(1+(1/n))^(1/2) ∼1+(1/(2n)) +(((1/2)((1/2)−1))/(2n^2 )) =1+(1/(2n))−(1/(8n^2 )) and (√(1−(1/n)))=1−(1/(2n))−(1/(8n^2 )) ⇒ (√(n+1))+(√(n−1))−2(√n)∼(√n)+(1/(2(√n)))−((√n)/(8n^2 )) +(√n)−(1/(2(√n)))−((√n)/(8n^2 )) −2(√n) =−((√n)/(4n^2 )) =−(1/4) n^((1/2)−2) =−(1/4)n^(−(3/2)) ⇒ U_n ∼n^(3/2) ×(−(1/4))n^(−(3/2)) ⇒U_n ∼−(1/4) ⇒ lim_(n→+∞) U_n =−(1/4)

$$\mathrm{let}\:\mathrm{U}_{\mathrm{n}} =\mathrm{n}^{\frac{\mathrm{3}}{\mathrm{2}}} \left\{\sqrt{\mathrm{n}+\mathrm{1}}+\sqrt{\mathrm{n}−\mathrm{1}}−\mathrm{2}\sqrt{\mathrm{n}}\right\}\:\Rightarrow \\ $$$$\mathrm{U}_{\mathrm{n}} =\mathrm{n}^{\frac{\mathrm{3}}{\mathrm{2}}} \left\{\sqrt{\mathrm{n}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}}+\sqrt{\mathrm{n}}\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{n}}}−\mathrm{2}\sqrt{\mathrm{n}}\right\}\:\mathrm{but} \\ $$$$\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}}=\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\sim\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2n}}\:+\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2n}^{\mathrm{2}} } \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2n}}−\frac{\mathrm{1}}{\mathrm{8n}^{\mathrm{2}} }\:\:\mathrm{and} \\ $$$$\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{n}}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2n}}−\frac{\mathrm{1}}{\mathrm{8n}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\sqrt{\mathrm{n}+\mathrm{1}}+\sqrt{\mathrm{n}−\mathrm{1}}−\mathrm{2}\sqrt{\mathrm{n}}\sim\sqrt{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{n}}}−\frac{\sqrt{\mathrm{n}}}{\mathrm{8n}^{\mathrm{2}} } \\ $$$$+\sqrt{\mathrm{n}}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{n}}}−\frac{\sqrt{\mathrm{n}}}{\mathrm{8n}^{\mathrm{2}} }\:−\mathrm{2}\sqrt{\mathrm{n}} \\ $$$$=−\frac{\sqrt{\mathrm{n}}}{\mathrm{4n}^{\mathrm{2}} }\:=−\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{n}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}} \:=−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{n}^{−\frac{\mathrm{3}}{\mathrm{2}}} \:\Rightarrow \\ $$$$\mathrm{U}_{\mathrm{n}} \sim\mathrm{n}^{\frac{\mathrm{3}}{\mathrm{2}}} ×\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)\mathrm{n}^{−\frac{\mathrm{3}}{\mathrm{2}}} \:\Rightarrow\mathrm{U}_{\mathrm{n}} \sim−\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{U}_{\mathrm{n}} =−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$

Commented by i jagooll last updated on 25/May/20

binomial series ?

$$\mathrm{binomial}\:\mathrm{series}\:? \\ $$

Commented by abdomathmax last updated on 25/May/20

no sir developpement of (1+x)^α at integr serie

$$\mathrm{no}\:\mathrm{sir}\:\mathrm{developpement}\:\mathrm{of}\:\left(\mathrm{1}+\mathrm{x}\right)^{\alpha} \:\mathrm{at}\:\mathrm{integr}\:\mathrm{serie} \\ $$

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