Prove that:

$$L=\lim_{n\to\infty} \frac {\sqrt 2 n^{\left(n-\frac 12\right)}}{n!}\left(\frac {(2\sqrt[n] {n} -1)^n}{n^2}\right)^{ \frac {n\left(n-\frac 12\right)}{\ln^2 n}}=\sqrt {\frac {e}{\pi}}$$

My method:

Properties I am going to use :

1)Stirling's approximation:$$n!\sim\sqrt {2\pi n} \left(\frac ne\right)^n$$

2)Property 2 : $$\sqrt[n] {n}\sim 1+\frac {\ln n}{n}+\frac{\ln^2 n}{2n^2}$$

3)Property 3: For all continuous and differentiable functions $f,g$ (In their domain respectively), if $\lim_{x\to\infty} g(x)=0$ then for large enough $x$ we have $$(1+g(x))^{f(x)}\sim e^{f(x)\cdot g(x)}$$

Using Stirling's approximation we get $$L=\lim_{n\to\infty} \frac {e^n}{\sqrt{\pi} n}\left(\frac {\displaystyle (2\sqrt[n]{n} -1)^n}{n^2}\right)^{\frac {n\left(n-\frac 12\right)}{\ln^2n}}$$

Using Property 2 we get $$L=\lim_{n\to\infty} \frac {e^n}{\sqrt {\pi} n}\left(\frac { \left(1+\frac {2\ln n}{n}+\frac{\ln^2n}{n^2}\right)^n}{n^2}\right)^{\frac {n\left(n-\frac 12\right)}{\ln^2n}}$$

And using the property 3 we get $$L=\lim_{n\to\infty} \frac {e^n}{\sqrt{\pi} n} \displaystyle \frac {e^{\frac {n(2n-1)}{\ln n}}\cdot e^{ \left(n-\frac 12\right)}}{ n^{\frac {n(2n-1)}{\ln^2 n}}}$$

Using that $$n^{\frac {n(2n-1)}{\ln^2 n}}=e^{\frac {n(2n-1)}{\ln n}}$$

Using this alongwith previous results we get $$L=\lim_{n\to\infty} \frac {e^n}{\sqrt{\pi} n} \displaystyle \frac {e^{\frac {n(2n-1)}{\ln n}}}{e^{ \frac {n(2n-1)}{\ln n}}}\cdot e^{ \left(n-\frac 12\right)}=\lim_{n\to\infty} \frac {e^{2n}}{\sqrt{e\pi} n}$$

Which clearly doesn't converge.Can someone please point out my mistake in above working. Also some new suggestions to solve this question will be quite beneficial.

  • $\begingroup$ The error arises when you substitute $1+\frac {\ln n}{n}+\frac{\ln^2 n}{2n^2}$ for $\sqrt[n] {n}$. While that would normally be a valid substitution, you are raising that to a power dependent on $n$, so the error gets "compounded". $\endgroup$ Apr 23, 2019 at 3:59
  • $\begingroup$ @automaticallyGenerated Ok Thanks, Can you please provide some suggestion to prove original limit $\endgroup$ Apr 23, 2019 at 4:06
  • $\begingroup$ Unless there is a very good reason, we prefer not to arbitrarily delete good content: the answers could well help other users in the future. $\endgroup$
    – robjohn
    Apr 23, 2019 at 8:04
  • $\begingroup$ @robjohn Oh sorry about that. Actually I solved it perfectly after automaticallyGenerated pointed out my mistake. So as the question was already solved by me, and at that moment there were no answers so I thought to delete the question. But since you have also answered the question ,its well and good now. I hope this question helps in any way to the future readers . Thanks for the advice.. $\endgroup$ Apr 23, 2019 at 8:17
  • $\begingroup$ It is perfectly fine to post answers for your own questions. In fact, that may help those who have looked at your question and want to know how to solve it. $\endgroup$
    – robjohn
    Apr 23, 2019 at 8:24

1 Answer 1


We can start with $$ \log\left(n^{1/n}\right)=\frac{\log(n)}n $$ and use the power series for $e^x$ to get $$ 2n^{1/n}-1=1+2\frac{\log(n)}n+\frac{\log(n)^2}{n^2}+\frac{\log(n)^3}{3n^3}+O\!\left(\frac{\log(n)^4}{n^4}\right) $$ The power series for $\log(1+x)$ yields $$ \begin{align} \log\left(2n^{1/n}-1\right) &=\overbrace{2\frac{\log(n)}n+\frac{\log(n)^2}{n^2}+\frac{\log(n)^3}{3n^3}}^x\overbrace{-2\frac{\log(n)^2}{n^2}-2\frac{\log(n)^3}{n^3}}^{-x^2/2}\overbrace{+\frac83\frac{\log(n)^3}{n^3}}^{x^3/3}\\ &=2\frac{\log(n)}n-\frac{\log(n)^2}{n^2}+\frac{\log(n)^3}{n^3}+O\!\left(\frac{\log(n)^4}{n^4}\right) \end{align} $$ Multiply by $n$ and use the power series for $e^x$ where $x=-\frac{\log(n)^2}n+\frac{\log(n)^3}{n^2}+O\!\left(\frac{\log(n)^4}{n^3}\right)$: $$ \left(2n^{1/n}-1\right)^n=n^2\left(1-\frac{\log(n)^2}n+\frac{\log(n)^4+2\log(n)^3}{2n^2}+O\!\left(\frac{\log(n)^6}{n^3}\right)\right) $$ Divide by $n^2$ and use the power series for $\log(1+x)$: $$ \log\left(\frac{\left(2n^{1/n}-1\right)^n}{n^2}\right)=-\frac{\log(n)^2}n+\frac{\log(n)^3}{n^2}+O\!\left(\frac{\log(n)^6}{n^3}\right) $$ Multiply by $\frac{n\left(n-\frac12\right)}{\log^2(n)}$, use the power series for $e^x$, and apply Stirling's Formula to get $$ \begin{align} \frac{\sqrt2n^{\left(n-\frac 12\right)}}{n!}\left(\frac{\left(2n^{1/n}-1\right)^n}{n^2}\right)^{\frac{n\left(n-\frac12\right)}{\log^2(n)}} &=\frac{\sqrt2n^{\left(n-\frac 12\right)}}{\sqrt{2\pi n}\,n^ne^{-n}}ne^{\frac12-n}\left(1+O\!\left(\frac{\log(n)^4}n\right)\right)\\ &=\sqrt{\frac e\pi}\left(1+O\!\left(\frac{\log(n)^4}n\right)\right) \end{align} $$ Therefore, $$ \lim_{n\to\infty}\frac{\sqrt2n^{\left(n-\frac 12\right)}}{n!}\left(\frac{\left(2n^{1/n}-1\right)^n}{n^2}\right)^{\frac{n\left(n-\frac12\right)}{\log^2(n)}}=\sqrt{\frac e\pi} $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .