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Question Number 8066 by Chantria last updated on 29/Sep/16

calculate lim_(n→∞) (n/2^(√n) )

$${calculate}\: \\ $$$$\:\underset{{n}\rightarrow\infty} {{lim}}\frac{{n}}{\mathrm{2}^{\sqrt{{n}}} } \\ $$$$ \\ $$

Answered by prakash jain last updated on 02/Oct/16

2^(√n) =e^((√n)ln 2) =1+(√n)ln 2+((((√n)ln 2)^2 )/(2!))+((((√n)ln 2)^3 )/(3!))+.. (n/2^(√n) )=(n/(1+(√n)ln 2+((((√n)ln 2)^2 )/(2!))+((((√n)ln 2)^3 )/(3!))+..)) =((n/n)/(1+((ln (√2))/n)+((ln 2)/(2!))+(((√n)ln 2)/(3!))+(higher power of n))) =(1/(1+((ln (√2))/n)+((ln 2)/(2!))+(((√n)ln 2)/(3!))+(higher power of n))) lim_(n→∞) (n/2^(√n) )=(1/(1+0+((ln 2)/(2!))+∞))=0

$$\mathrm{2}^{\sqrt{{n}}} ={e}^{\sqrt{{n}}\mathrm{ln}\:\mathrm{2}} =\mathrm{1}+\sqrt{{n}}\mathrm{ln}\:\mathrm{2}+\frac{\left(\sqrt{{n}}\mathrm{ln}\:\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2}!}+\frac{\left(\sqrt{{n}}\mathrm{ln}\:\mathrm{2}\right)^{\mathrm{3}} }{\mathrm{3}!}+.. \\ $$$$\frac{{n}}{\mathrm{2}^{\sqrt{{n}}} }=\frac{{n}}{\mathrm{1}+\sqrt{{n}}\mathrm{ln}\:\mathrm{2}+\frac{\left(\sqrt{{n}}\mathrm{ln}\:\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2}!}+\frac{\left(\sqrt{{n}}\mathrm{ln}\:\mathrm{2}\right)^{\mathrm{3}} }{\mathrm{3}!}+..} \\ $$$$=\frac{\frac{{n}}{{n}}}{\mathrm{1}+\frac{\mathrm{ln}\:\sqrt{\mathrm{2}}}{{n}}+\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}!}+\frac{\sqrt{\mathrm{n}}\mathrm{ln}\:\mathrm{2}}{\mathrm{3}!}+\left(\mathrm{higher}\:\mathrm{power}\:\mathrm{of}\:{n}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{ln}\:\sqrt{\mathrm{2}}}{{n}}+\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}!}+\frac{\sqrt{\mathrm{n}}\mathrm{ln}\:\mathrm{2}}{\mathrm{3}!}+\left(\mathrm{higher}\:\mathrm{power}\:\mathrm{of}\:{n}\right)} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}}{\mathrm{2}^{\sqrt{{n}}} }=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{0}+\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}!}+\infty}=\mathrm{0} \\ $$

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