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I can show that $\displaystyle a_n=\left(n\cdot \ln\left(\frac{n+1}{n}\right)\right)^n\rightarrow \frac{1}{\sqrt{e}}$ by expressing it as $\displaystyle e^{\ln(a_n)}$, but this ends up very tedious. What is an easier way to compute this limit?

Thanks!

Edit This is a sequence, so I mean the limit as $\displaystyle n\rightarrow \infty$.

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As $n$ goes to what? –  FUZxxl Feb 26 '12 at 19:58
    
Infinity (it's a sequence) –  josh Feb 26 '12 at 20:10
    
@J.D. I have removed the asymptotics tags, as the question just asks for the limit. –  Aryabhata Feb 26 '12 at 20:17
    
@Aryabhata my mistake. I tagged based on one of the answers not based on the question. Thanks for the feedback. –  user2468 Feb 26 '12 at 20:19
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3 Answers

up vote 4 down vote accepted

Asymptotics

$$\begin{align} a_n &= \left(n\cdot \ln\left(\frac{n+1}{n}\right)\right)^n = \left(n\cdot \ln\left(1+\frac{1}{n}\right)\right)^n = \left(n\cdot \left(\frac{1}{n}-\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)\right)\right)^n \\ &= \left(1-\frac{1}{2n}+o\left(\frac{1}{n}\right)\right)^n = \left(\left(1-\frac{1}{n}+o\left(\frac{1}{n}\right)\right)^{1/2}\right)^{n} = \left(\left(1-\frac{1}{n}+o\left(\frac{1}{n}\right)\right)^n\right)^{1/2} \\ &= \left(\left(e^{-1/n}+o\left(\frac{1}{n}\right)\right)^n\right)^{1/2} = (e^{-1}+o(1))^{1/2} =e^{-1/2}+o(1) . \end{align}$$

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We can simplify the proof a little if we use: If $c_n \to c$, then $(1 + \frac{c_n}{n})^n \to e^c$. Now $1 - \frac{1}{2n} + o(\frac{1}{n}) = 1 + \frac{c_n}{n}$ where $c_n \to -\frac{1}{2}$. –  Aryabhata Feb 26 '12 at 20:19
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Or a little bit more explicit, based on the Taylor expansion of $\log$, the following inequalities hold for $n \geq 2$:

$$ \left(1-\frac{1}{2n}\right)^n \leq a_n \leq \left(1-\frac{1}{2n}+\frac{1}{3n^2}\right)^n \leq \left(1-\frac{1}{2(n+1)}\right)^n $$

and both sides have limit $e^{-1/2}$.

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Use \left( and \right) to get this $$\left(\frac{\log x}{1-x^2}\right)$$ instead of this $$(\frac{\log x}{1-x^2})$$ –  Pedro Tamaroff Mar 6 '12 at 6:51
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You can also use the Trapezoidal rule to approximate $$ \ln \left(\frac{n+1}{n} \right)=\int_n^{n+1}{\frac{1}{x}}dx=\frac{1}{2}\left(\frac{1}{n+1}+\frac{1}{n}\right)+\mathcal{O}\left(\frac{1}{n^3}\right) $$ Exponentiating and multiplying by $n^n$ yields $$ n^n \ln \left(\frac{n+1}{n}\right)^n=\left[\left(\frac{2n+1}{2n+2}\right)+\mathcal{O}\left(\frac{1}{n^2}\right)\right]^n=\left(1-\frac{1}{2n}+\mathcal{o}\left(\frac{1}{n}\right)\right)^n=e^{-\frac{1}{2}}+\mathcal{o}(1) $$ (The last equal sign is already given in the answer of GEdgar).

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Use \left( and \right) to get this $$\left(\frac{x-1}{x^2+1}\right)$$ instead of this $$(\frac{x-1}{x^2+1})$$ Also use a slash \ before function names like: \log, \sin, \exp, to get $\log$, $\sin$, $\exp$ instead of $log$, $sin$, $exp$ –  Pedro Tamaroff Mar 6 '12 at 6:49
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