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When I was trying to solve a problem to find the radius of convergence of the power series

$$\sum \frac{2^nz^n}{n!}$$

I fully understand that the ratio test works well in this one and the radius of convergence is $\infty$.

However, knowing that the root test gives a better span of the ratio test and it is necessary to prove it, I wanted to be able to find the radius of convergence using the ratio test.

Thus obtaining the following

$$\begin{align} \lim_{n \to \infty} \sqrt[n]{\frac{2^nz^n}{n!}} & = |2z|\lim_{n \to \infty} \sqrt[n]{\frac{1}{n!}} \\ \\ & = 0\\ \\ & \lt 1 \\ \\ & \Rightarrow R = \infty \end{align}$$

must be true.

So, I was thinking that $\lim_{n \to \infty} \sqrt[n]{\frac{1}{n!}}$ must be equal to 0.

Is there a direct proof of this ?

I just want to be a bit more algebraically savvy.

Américo Tavares
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hyg17
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5 Answers5

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$$n! >n(n-1)..(n-\lfloor \frac{n-1}{2} \rfloor)> (\frac{n}{2})^{\frac{n}{2}}$$

Thus

$$\sqrt[n]{\frac{1}{n!}}< \frac{\sqrt2}{\sqrt{n}}$$

N. S.
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  • Could you explain me the first inequality ? I am not familiar with it. I tried to verify it, too, only to fail. – hyg17 Jun 05 '13 at 05:44
  • @hyg17: the first inequality follows directly from the definition of $n!$ – Simone Jun 05 '13 at 11:39
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You can use Stirling's approximation: $\displaystyle n!\approx\sqrt{2\pi n}\bigg(\frac{n}{e}\bigg)^n$ as $\displaystyle n\to\infty$, so that $\displaystyle\bigg(\frac{1}{n!}\bigg)^{\frac{1}{n}}\approx \bigg(\frac{1}{\sqrt{2\pi n}\big(\frac{n}{e}\big)^n}\bigg)^{\frac{1}{n}}=\bigg(\frac{1}{(2\pi n)^{\frac{1}{2n}}\big(\frac{n}{e}\big)}\bigg)$ which clearly goes to zero as $n\to\infty$.

JLA
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  • Wow, I've only seen this on a text book on statistics, and I am not familiar with it at all. But thanks for your help. – hyg17 Jun 05 '13 at 05:46
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Are you aware of the inequality:

$$\liminf_{n \to \infty} \frac{a_{n+1}}{a_n} \le \liminf_{n \to \infty} a_n^{\frac{1}{n}} \le \limsup_{n \to \infty} a_n^{\frac{1}{n}} \le \limsup_{n \to \infty} \frac{a_{n+1}}{a_n}$$

Now just notice that if $a_n = \frac{1}{n!}$, then $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{1}{n+1}$$

A.S
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  • I'm not! Any hints to show it? It looks nice on its own, regardless of OP's question. – Patrick Da Silva Jun 04 '13 at 19:35
  • @PatrickDaSilva Proving the inequalities with limsup and liminf are very similar. Let $\beta > \limsup \frac{a_{n+1}}{a_n}$ and prove that $a_{n+p} \le \beta^p a_n$ for sufficiently large $n$. Take the $n+p$-th root of both sides and let $p \to \infty$. You can also see $3.37$ in Baby Rudin or search "root test is stronger than ratio test." – A.S Jun 04 '13 at 19:47
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    @PatrickDaSilva Here it is – Pedro Jun 04 '13 at 19:58
  • That actually makes sense, because this is exactly why we know the ratio test is derived from the root test. This is very helpful, thank you . – hyg17 Jun 05 '13 at 05:45
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You may use the fact that $n! \geq e(n/e)^{n},\,n \geq 1$.

Lord Soth
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  • Is this a famous inequality ? If so I would appreciate it if you could tell me where to go to learn more about this. – hyg17 Jun 05 '13 at 05:47
  • This comes from approximating a sum via an integral (I do not think it is particularly famous), i.e. $\int_1^n \log x dx \leq \sum_{x=1}^n \log x$. The LHS is $n\log(n/e)+1$, the RHS is $\log n!$. Now take the exponents of both sides. I found it on Wikipedia. – Lord Soth Jun 05 '13 at 18:38
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We can apply the following proposition to $u_{n}=\dfrac{1}{n!}$, a proof of which can be found in this post of mine in Portuguese. I translate it below.

Proposition. Assume that for all $n$, $u_{n}>0$ and $\lim_{n\rightarrow \infty }\dfrac{u_{n+1}}{u_{n}}=b$. Then $$\lim_{n\rightarrow \infty }\sqrt[n]{u_{n}}=b.$$

Proof. $\lim_{n\rightarrow \infty }\dfrac{u_{n+1}}{u_{n}}=b$ implies that there exists a natural number $N$ such that for $n\ge N$ we have $$ \begin{equation*} b-\delta <\frac{u_{N+k+1}}{u_{N+k}}<b+\delta ,\qquad 0\leq k\leq n-N-1. \end{equation*} $$

Multiplying these $n-N$ inequalities, we get $$ \begin{eqnarray*} \left( b-\delta \right) ^{n-N} &=&\prod_{k=0}^{n-N-1}\left( b-\delta \right) <\prod_{k=0}^{n-N-1}\frac{u_{N+k+1}}{u_{N+k}}<\prod_{k=0}^{n-N-1}\left( b+\delta \right) =\left( b+\delta \right) ^{n-N}\end{eqnarray*} $$

Since the product $\displaystyle\prod_{k=0}^{n-N-1}\dfrac{u_{N+k+1}}{u_{N+k}}=\dfrac{u_{n}}{u_{N}}$, we thus have

$$ \begin{eqnarray*} \left( b-\delta \right) ^{n-N} &<&\frac{u_{n}}{u_{N}}<\left( b+\delta \right) ^{n-N}\end{eqnarray*} $$

Multiplying by $u_N$, we get

$$ \begin{eqnarray*} \left( b-\delta \right) ^{n-N}u_{N} &<&u_{n}<\left( b+\delta \right) ^{n-N}u_{N}. \end{eqnarray*} $$

Hence $$ \begin{equation*} \left( b-\delta \right) \sqrt[n]{\left( b-\delta \right) ^{-N}u_{N}}<\sqrt[n]{u_{n}}<\left( b+\delta \right) \sqrt[n]{\left( b+\delta \right) ^{-N}u_{N}}. \end{equation*} $$

Since $\lim_{n\to \infty}\sqrt[n]{\left( b+\delta \right) ^{-N}u_{N}}=1$, there exists a natural number $N^{\prime }$ such that for $n\geq N^{\prime }$, $$1-\delta<\sqrt[n]{\left( b-\delta \right) ^{-N}u_{N}}<1+\delta$$

which means that

$$ \begin{equation*} b-\delta -b\delta +\delta ^{2}=(1-\delta )\left( b-\delta \right) <\sqrt[n]{u_{n}}<\left( b+\delta \right) (1+\delta )=b+b\delta +\delta +\delta ^{2}. \end{equation*} $$

For $\varepsilon =\delta +b\delta +\delta ^{2}$ and $n\geq \max \{N,N^{\prime }\}$ $$ \begin{equation*} b-\varepsilon <\sqrt[n]{u_{n}}<b+\varepsilon , \end{equation*} $$ thus proving the proposition. $\qquad\square$

For $u_{n}=\dfrac{1}{n!}$, we have $$ \begin{equation*} \lim_{n\rightarrow \infty }\frac{u_{n+1}}{u_{n}}=\lim_{n\rightarrow \infty } \frac{1/(n+1)!}{1/n!}=\lim_{n\rightarrow \infty }\frac{1}{n+1}=0. \end{equation*} $$

Consequently, $$ \begin{equation*} \lim_{n\rightarrow \infty }\sqrt[n]{u_{n}}=\lim_{n\rightarrow \infty }\sqrt[n]{1/n!}=0. \end{equation*} $$

Américo Tavares
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