5
$\begingroup$

Question is to evaluate :

$$\lim_{n\to \infty}\left\{ \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\dots\left(1+\frac{n}{n}\right)\right\}^{\frac{1}{n}}$$

I tried to do something like this but it is not getting better...

$(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{n}{n})=(\frac{n+1}{n})(\frac{n+2}{n})\cdots (\frac{n+n}{n})=\frac{1}{n^n}(n+1)(n+2)\dots(n+n)$

please give some hint to proceed further.

$\endgroup$
2
  • $\begingroup$ Please, try to make the title of your questions more informative. E.g., Why does $a\le b$ imply $a+c\le b+c$? is much more useful for other users than A question about inequality. For more information on choosing a good title, see this post. $\endgroup$
    – Lord_Farin
    Nov 5 '13 at 10:58
  • $\begingroup$ Thanks for your edit.. I would keep that in mind... $\endgroup$
    – user87543
    Nov 5 '13 at 11:01
10
$\begingroup$

Consider the log of $[(1+\frac{1}{n})(1+\frac{2}{n})\dots(1+\frac{n}{n})]^{\frac{1}{n}}$:

$$\frac{1}{n}\sum_{k=1}^n \ln \left( 1+\frac{k}{n}\right).$$

The limit of this sum is equal to $\int_0^1\ln(1+x)dx$, so

$$ \begin{aligned} \lim_{n\to\infty} [(1+\frac{1}{n})(1+\frac{2}{n})\dots(1+\frac{n}{n})]^{\frac{1}{n}} &= \lim_{n\to\infty} \exp\left(\frac{1}{n}\sum_{k=1}^n \ln \left( 1+\frac{k}{n}\right) \right)\\ &= \exp\left(\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n \ln \left( 1+\frac{k}{n}\right) \right)\\ &= \exp \left( \int_0^1\ln(1+x)dx\right) \end{aligned}$$

$\endgroup$
5
  • 1
    $\begingroup$ I have tried to complete that.. please see if it is correct... $\int_0^1\ln(1+x)dx=(1+x)\ln(1+x)-(1+x)|^{x=1}_{x=0}=2\ln(2)-2+1=2\ln(2)-1$ and $\exp (\int_0^1\ln(1+x)dx)=e^{2\ln(2)-1}=e^{\ln(4)-1}=\frac{4}{e}$ $\endgroup$
    – user87543
    Nov 5 '13 at 9:33
  • $\begingroup$ only thing i did not understood is how did you get last step i mean from limit of sum to integral sign... I can some how convince my self that sum became limit ans as $k$ varies over all natural numbers we get $x$ but where does $\frac{1}{n}$ gone :O $\endgroup$
    – user87543
    Nov 5 '13 at 9:43
  • $\begingroup$ @PraphullaKoushik If you're familiar with the definition of integral as a Riemann sum... $\endgroup$ Nov 5 '13 at 10:10
  • $\begingroup$ @CarlosEugenioThompsonPinzón : Yes.. It strikes it should be the case but i did not realize the proof immediately... i would spend some time on this.. Thank You :) $\endgroup$
    – user87543
    Nov 5 '13 at 10:15
  • $\begingroup$ @praphulla: You can imagine the $1/n$ factor as base-line of a rectangle of height $ \log(1+k/n) $ and then recall, that integration can be seen as limit of computation of the area below a curve ("quadrature"), when separated in $n$ parts of $1/n$-width... $\endgroup$ Nov 5 '13 at 10:41
9
$\begingroup$

$$(\frac{n+1}{n})(\frac{n+2}{n})\dots (\frac{n+n}{n})=\frac{1}{n^n}(n+1)(n+2)\dots(n+n)$$

$$=\lim_{n\rightarrow \infty}\left[\frac{1}{n^n}\frac{(2n)!}{n!}\right]^{1/n}=\lim_{n\rightarrow \infty}\left[\frac{\sqrt{4\pi n}\cdot 2^{2n}\frac{n^{2n}}{e^{2n}}}{\sqrt{2\pi n}\cdot \frac{n^{2n}}{e^n}}\right]^{1/n}=\lim_{n\rightarrow \infty }[\sqrt{2}]^{1/n}\cdot \frac{4}{e}=\frac{4}{e}$$

Here I've used the Stirling approximation of $n!$,i.e- $n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}$

$\endgroup$
3
  • $\begingroup$ Thanks for your answer...I am not familiar with Stirling approximation but i am sure it will be useful in future,,, thank You for spending time on my question.. $\endgroup$
    – user87543
    Nov 5 '13 at 9:58
  • $\begingroup$ In your last expression there is a typo: It should be $\large n$ instead of $\large 1/n$ in the exponent. $\endgroup$ Nov 5 '13 at 10:25
  • 1
    $\begingroup$ @shaswata. When n is large, Log(n!) is very well approximated by n Log[n] - n. For sure, this gives the same result ! $\endgroup$ Nov 5 '13 at 11:06

Your Answer

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