Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
Showing that $\frac{\sqrt[n]{n!}}{n}$ $\rightarrow \frac{1}{e}$

Any suggestions to find the following limit: $$\displaystyle\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{\frac{1}{n}}$$ with basic tools of the calculus!

share|cite|improve this question

marked as duplicate by Marvis, David Mitra, amWhy, Henry T. Horton, Argon Jan 8 '13 at 1:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

What exactly do you mean by basic tools of the calculus? – Andrew D Jan 8 '13 at 1:06
something like the convergence tests; something easy for a first year undergraduates. – Juan Carlos Castro Jan 8 '13 at 1:10
See the proof of Theorem 1 in this document (PDF). – Antonio Vargas Jan 8 '13 at 1:11
If you can use this, you're fine:'s_approximation. But I am not sure this qualifies as a basic tool. – 1015 Jan 8 '13 at 1:16
Duplicate: – Jack Jan 8 '13 at 1:26

2 Answers 2

Below is a quick derivation of part of Stirling's formula, which will enable you to conclude that the limit is $1/e$. First note that from Abel summation, we have that $$\sum_{n=1}^N a(n) b(n) = \sum_{n=1}^N b(n)(A(n)-A(n-1)) = \sum_{n=1}^{N} b(n) A(n) - \sum_{n=1}^N b(n)A(n-1)\\ = \sum_{n=1}^{N} b(n) A(n) - \sum_{n=0}^{N-1}^N b(n+1)A(n) = b(N) A(N) - b(1)A(0) + \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$ where $A(n) = \displaystyle \sum_{k=1}^n a(k)$.

In our case, take $a(n) = 1$, $b(n) = \log(n)$. This gives us $A(n) = n$. \begin{align} \log(n!) = \sum_{k=1}^n \log(k) & = \sum_{k=1}^n a(k) b(k) = n \log(n) + \sum_{k=1}^{n-1}k (\log(k) - \log(k+1))\\ & = n \log(n) + \sum_{k=1}^{n-1}k \log \left(1 - \dfrac1{k+1} \right)\\ & = n \log(n) + \sum_{k=1}^{n-1}k \left(-\dfrac1{k+1} + \mathcal{O}\left(\dfrac1{(k+1)^2} \right)\right)\\ & = n \log(n) + \sum_{k=1}^{n-1} \left(-\dfrac{k}{k+1} + \mathcal{O}\left(\dfrac{k}{(k+1)^2} \right)\right)\\ & = n \log(n) + \sum_{k=1}^{n-1} \left(-1 + \mathcal{O}\left(\dfrac1{(k+1)} \right)\right)\\ & = n \log(n) - n + \mathcal{O}\left(\log(n) \right)\\ \end{align} Hence, we get that $$\log \left(\dfrac{n!}{n^n}\right) = - n + \mathcal{O}(\log (n))$$ Hence, $$\dfrac1n \log \left(\dfrac{n!}{n^n}\right) = - 1 + \mathcal{O}\left(\dfrac{\log (n)}n \right)$$ Hence, we get that $$\lim_{n \to \infty} \dfrac1n \log \left(\dfrac{n!}{n^n}\right) = -1$$ And since $\log$ is continuous, we get that $$\lim_{n \to \infty}\left(\dfrac{n!}{n^n}\right)^{1/n} = \dfrac1e$$

share|cite|improve this answer
MSE should award you the ‘Gold Badge for Abel Summation’! – Haskell Curry Jan 8 '13 at 1:47
@HaskellCurry It is one thing which I know and hence I tend to over use it :). – user17762 Jan 8 '13 at 2:24

I could not find the following proof in the multiple duplicates, so here it is.

By integral comparison, $\int_{k-1}^k\ln t dt\leq \ln k \leq \int_k^n \ln t dt$ for all $k=2,\ldots,n$. Summing these $n-1$ inequations yields: $$ \int_1^n\ln t dt \leq \ln (n!) \leq \int_2^{n+1}\ln t dt. $$ Using that $t\ln t-t$ is an antiderivative of $\ln t$, we get: $$ n\ln n-n+1\leq \ln (n!) \leq (n+1)\ln(n+1)-n+C $$ where $C=1-2\ln 2$.

Next $$ -n+1\leq \ln (n!) - n\ln n\leq n\ln(1+1/n) + \ln(n+1) -n+C $$ whence $$ -1+\frac{1}{n} \leq \frac{ \ln (n!) - n\ln n}{n} \leq -1 + \ln(1+1/n) +\frac{\ln(n+1)+C}{n}. $$

By the Squeeze Theorem, it follows that $$ \lim_{n\rightarrow +\infty}\frac{ \ln (n!) - n\ln n}{n}=-1 $$ and finally $$ \lim_{n\rightarrow +\infty} \left( \frac{n!}{n^n}\right)^{1/n}=\lim_{n\rightarrow +\infty} \exp\left( \frac{ \ln (n!) - n\ln n}{n} \right)=\exp(-1). $$

share|cite|improve this answer

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