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?


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.