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I don't know how to prove that $$\lim_{n\to\infty} \frac{n}{\sqrt[n]{n!}} =e.$$ Are there other different (nontrivial) nice limit that gives $e$ apart from this and the following $$\sum_{k = 0}^\infty \frac{1}{k!} = \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = e\;?$$

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Do you want a proof for the first limit ? –  Amr Mar 5 '13 at 22:27
    
The bottom right one, with the infinite limit, is the original. That is the ACTUAL definition. When Bernoulli originally came up with the concept. Every other equality you come across had to have been proven. –  CogitoErgoCogitoSum Mar 5 '13 at 22:53
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@Cogito, it is one possible definition. You can take any one of a number of formulas as the definition, then prove the others from the one you've chosen. –  Gerry Myerson Mar 5 '13 at 23:19
    
Almost the same as this one: math.stackexchange.com/q/201906/9464 –  Jack Mar 6 '13 at 0:18
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6 Answers

up vote 6 down vote accepted

In the series for $$e^n=\sum_{k=0}^\infty \frac{n^k}{k!},$$ the $n$th and biggest(!) of the (throughout positve) summands is $\frac{n^n}{n!}$. On the other hand, all summands can be esimated as $$ \frac{n^k}{k!}\le \frac{n^n}{n!}$$ and especially those with $k\ge 2n$ can be estimated $$ \frac{n^k}{k!}<\frac{n^{k}}{(2n)^{k-2n}\cdot n^{n}\cdot n!}=\frac{n^{n}}{n!}\cdot \frac1{2^{k-2n}}$$ and thus we find $$\begin{align}\frac{n^n}{n!}<e^n&=\sum_{k=0}^{2n}\frac{n^k}{k!}+ \sum_{k=2n}^\infty \frac{n^k}{k!}\\&<(2n+1)\cdot\frac{n^n}{n!}+ \frac{n^n}{n!}\sum_{k=0}^\infty 2^{-k}\\&=(2n+3)\cdot\frac{n^n}{n!}.\end{align}$$ Taking $n$th roots we find $$ \frac n{\sqrt[n]{n!}}\le e\le \sqrt[n]{2n+3}\cdot\frac n{\sqrt[n]{n!}}.$$ Because $\sqrt[n]{2n+3}\to 1$ as $n\to \infty$, we obtain $$\lim_{n\to\infty}\frac n{\sqrt[n]{n!}}=e$$ from squeezing.

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There are as many representations of $e$ as you could want at Wikipedia. $$e=\sum_1^{\infty}{k^7\over877k!}$$ $$e=\lim_{n\to\infty}n^{\pi(n)/n}$$ where $\pi(n)$ is the number of primes up to $n$. Any many, many more.

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A corollary of your answer seems to be that one can want only a finite number of representations of $e$. –  Lepidopterist Mar 5 '13 at 22:49
    
$\sum_k{k^n\over k!B_n}$ is there, and gives a countable infinity of representations, one for each non-negative integer $n$ (the $B_n$ are the Bell numbers). –  Gerry Myerson Mar 5 '13 at 23:16
    
So one can only wish for countable representations, you're saying? –  Lepidopterist Mar 5 '13 at 23:19
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I will use Cauchy-D'Alembert criterion

$$\lim_{n\to\infty} \frac{n}{\sqrt[n]{n!}} =e$$

you can write the limit as $$\sqrt[n]{\frac{n^{n}}{n!}}.$$

Let be $\displaystyle x_{n}=\frac{n^{n}}{n!}.$

Now we can do $$\frac{x_{n+1}}{x_{n}}=\frac{\frac{(n+1)^{n}\cdot (n+1)}{n! \cdot (n+1)}}{\frac{n^n}{n!}}=\frac{(n+1)^{n}\cdot (n+1)}{n! \cdot (n+1)}\cdot \frac{n!}{n^{n}}=\frac{(n+1)^{n}}{n^n}=\left(\frac{n+1}{n}\right)^{n}=\left(1+\frac{1}{n}\right)^n \to e. $$

So $\displaystyle \lim_{n\to\infty} \frac{n}{\sqrt[n]{n!}} =e.$

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Since $$ \ln \left( \frac{n}{\sqrt[n]{n!}}\right)=\ln n-\frac{\ln n!}{n} $$ all you need is the weak Stirling formula: $$ \ln n!=n\ln n -n +O(\ln n). $$ By comparison with the integral of the nondecreasing function $\ln t$: $$ \int_1^n\ln tdt\leq \ln n! =\sum_{k=2}^n \ln k\leq \int_2^{n+1}\ln t dt. $$ Recall that $\int \ln tdt =t\ln t -t +C$ and compute the lhs and the rhs.

The weak Stirling formula follows easily.

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Using stirling's approximation seems to be the most succinct way of showing this. Nice, +1 –  Rustyn Mar 5 '13 at 23:54
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Use Stirling's approximation

$$\frac{n}{\sqrt[n]{n!}}\cong\frac{n}{\sqrt[n]{\sqrt{2\pi }\,n^{n+1/2}e^{-n}}}=\frac{1}{(2\pi)^{1/2n}}\frac{e}{n^{1/2n}}\xrightarrow[n\to\infty]{}1\cdot\frac{e}{1}=e$$

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I found this here:

$$e = \lim_{n \to \infty} \sqrt[n]{n\#}$$

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