Infinite Series

Real Analysis
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jacks
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Infinite Series

#1

Post by jacks »

Calculation of \(\displaystyle \lim_{n\rightarrow \infty}\prod_{r=1}^{n} \left(1+\frac{r}{n}\right)^{\frac{1}{n}}\)

Can we solve the above limit using Stirling approximation OR by using Stolz-Cesaro Theorem.

Thanks
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Tolaso J Kos
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Re: Infinite Series

#2

Post by Tolaso J Kos »

So here goes. We are using the obvious identity of: \( \displaystyle \ln \left ( \prod x_i \right )=\sum \ln x_i \).
Now, let \( L_n \) denote the given limit then: $$L_n=\prod_{r=1}^{n}\left ( 1+\frac{r}{n} \right )^{1/n} \implies \ln L=\ln \left [ \prod_{r=1}^{n}\left ( 1+\frac{r}{n} \right )^{1/n} \right ]=\frac{1}{n}\sum_{r=1}^{n}\ln \left ( 1+\frac{r}{n} \right )$$ Taking limit on both sides we see that as \( n \to +\infty \) the right side is nothing else than a \( \rm{Riemmann} \) sum of the \( \ln (1+x) \) function. Then: \( \displaystyle \ln L_n=\int_{0}^{1}\ln \left ( 1+x \right )\,dx = \ln 4-1 \) .

So: $$L_n=\frac{4}{e}$$ which evaluates the given limit and we're done.
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Re: Infinite Series

#3

Post by Demetres »

To use Stirling's approximation observe that \[ \prod_{r=1}^n \left(1 + \frac{r}{n}\right)^{1/n} = \left( \frac{(n+1)(n+2)\cdots (n+n)}{n^n}\right)^{1/n} = \left(\frac{(2n)!}{n^n(n!)}\right)^{1/n}.\] By Stirling's approximation we have \[ \frac{(2n)!}{n^n(n!)} \sim \frac{\sqrt{4\pi n}(2n/e)^{2n}}{n^n \sqrt{2\pi n}(n/e)^n} = \sqrt{2}(4/e)^n.\] It is now simple to get that the required limit is \(4/e\).
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