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Question Number 197947 by mnjuly1970 last updated on 05/Oct/23

            calculate…    L = lim _(n→∞) (( (1+(1/2) )(1+(1/3))… (1+(1/n))))^(1/n)  = ?

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{calculate}\ldots \\ $$$$\:\:\mathrm{L}\:=\:\mathrm{lim}\:_{\mathrm{n}\rightarrow\infty} \sqrt[{{n}}]{\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)\ldots\:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)}\:=\:?\:\:\:\:\:\:\:\: \\ $$$$\:\: \\ $$

Answered by MM42 last updated on 05/Oct/23

A=lim_(n→∞)  ((3/2)×(4/3)×...×(n/(n−1))×((n+1)/n))^(1/n)   lnA=lim_(n→∞)  ln(((n+1)/2))^(1/n)     =lim_(n→∞)  ((ln(((n+1)/2)))/n)=0⇒A=1 ✓

$${A}={lim}_{{n}\rightarrow\infty} \:\left(\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{3}}×...×\frac{{n}}{{n}−\mathrm{1}}×\frac{{n}+\mathrm{1}}{{n}}\right)^{\frac{\mathrm{1}}{{n}}} \\ $$$${lnA}={lim}_{{n}\rightarrow\infty} \:{ln}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{{n}}} \\ $$$$ \\ $$$$={lim}_{{n}\rightarrow\infty} \:\frac{{ln}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)}{{n}}=\mathrm{0}\Rightarrow{A}=\mathrm{1}\:\checkmark \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 05/Oct/23

thx alot sir ...

$${thx}\:{alot}\:{sir}\:... \\ $$

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