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Please note that this is homework. Please excuse my lack of $\LaTeX{}$ knowledge.

The Problem:

Evaluate the given limit by first recognizing the sum (possibly after taking the logarithm to transform the product into a sum) as a Riemann sum of appropriate function associated to a regular partition of $[0,1]$.

$\lim_{n\to \infty}\frac{1}{n}\sqrt[n]\frac{(2n)!}{n!}$

My petty Attempt

Okay so taking the hint from the problem. I tried this:

$\Delta x = \dfrac{1-0}{n} = \frac{1}{n}$

Then $x^*_k = 0 + k\Delta x = \frac{k}{n}$

Then I took the logarithm and expanded the logarithm.

$e^{[\lim_{n\to \infty}ln{(\frac{1}{n}\sqrt[n]\frac{(2n)!}{n!})}]}$

$e^{[\lim_{n\to \infty}ln{(\frac{1}{n})} + ln{(\sqrt[n]\frac{(2n)!}{n!})}]}$

$e^{[\lim_{n\to \infty}-ln{(n)} + \frac{ln(\frac{(2n)!}{n!})}{n}]}$

$e^{[\lim_{n\to \infty}-ln{(n)} + \frac{ln(2n!) - ln(n!)}{n}]}$

Now after this I am totally stuck. I am not even sure if I am going on the right track. All I ask is for some guidance. I know that the Riemann Sum is defined as:

$\sum^n_{k=1}{f(x^*_k)\Delta x_k}$

I also understand that the stuff inside the limit is already in closed form. Somehow I have to use the properties of logarithm to transform it to a $\sum$ but how?

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We apply the $\log$ function and we use the Riemann sum and we integrate: $$\log\left(\frac{1}{n}\sqrt[n]\frac{(2n)!}{n!}\right)=-\log n+\frac{1}{n}\sum_{k=1}^n\log(k+n)=\frac{1}{n}\sum_{k=1}^n\log(k+n)-\log n\\=\frac{1}{n}\sum_{k=1}^n\log\left(\frac{k}{n}+1\right)\to\int_0^1 \log(1+x)dx=2\log(2)-1$$

hence $$\lim_{n\to\infty}\frac{1}{n}\sqrt[n]\frac{(2n)!}{n!}=\frac{4}{e}$$

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  • $\begingroup$ Is the answer to that integral (I assume it's the natural log) the answer to the limit problem? I think we have to raise it back to the e-power? $\endgroup$ – imranfat May 28 '13 at 16:55
  • $\begingroup$ OK, makes sense now.... $\endgroup$ – imranfat May 28 '13 at 16:58

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