# How to prove that $n^2(2^n) / n! \to 0$ as $n \to \infty$

I'm trying to see how I can prove that :

$$\frac{n^22^n}{n!}\xrightarrow [n\to\infty]{} 0$$

Not sure how to show this..anyone care to explain? Thanks!

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– Peter Phipps Feb 20 '13 at 20:33

A fancy way: check for convergence/divergence of the series $\,\displaystyle{\sum_{n=1}^\infty\frac{n^22^n}{n!}}\,$ , for example with the ratio test (D'Alembert's test):

$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^22^{n+1}}{(n+1)!}\frac{n!}{n^22^n}=2\frac{n+1}{n^2}\xrightarrow[n\to\infty]{} 0<1$$

Thus, the ratio test tells us the series converges and so its general term sequence converges to zero.

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Actually, D'Alambert's criterion simply tells that $a_n<q^n$ for $q<1$ which is enough to tell the sequence goes to $0$. The use for series is a "plus", since we know $q^n,q<1$ is summable. – Pedro Tamaroff Feb 20 '13 at 20:00
I think Peter was just pointing out that $a_{n+1}/a_n\rightarrow q<1$ alone will imply $(a_n)$ has limit $0$. There is no need to talk about series. – David Mitra Feb 20 '13 at 20:07
Oh, perhaps that's right, @David Mitra, yet as he mentioned the "criterion" I thought he referred to the series thing. – DonAntonio Feb 20 '13 at 20:11
Yes, that is what I meant @DavidMitra – Pedro Tamaroff Feb 20 '13 at 20:15

Here's a hint: \begin{align} \frac{n^2 2^n}{n!}&=\frac{\overbrace{2\times 2 \times \cdots \times 2}^{n \mbox{ times}} \times n^2}{1\times 2\times \cdots \times 3\times n}\\ &=\frac{n}{n-1} \times 4 \cdot \frac{\overbrace{2\times 2\times \cdots \times 2}^{(n-2) \mbox{ times}}}{1\times 2\times \cdots \times (n-2)}\\ \end{align} If you still need help, please indicate so in the comments.

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Hint for the hint: $={4n\over n-1}\cdot{2\over 3}\cdot2\cdot \Bigl({{2\over n-2}\cdot{2\over n-3}\cdots{2\over4}}\Bigr)$. – David Mitra Feb 20 '13 at 20:16

By Stirling's Approximation:

$$\frac{n^2 2^n}{n!} \to \frac{(2e)^n}{\sqrt{2\pi}n^{n-1.5}}\sim \frac{1}{(n/2e)^n}\to 0$$

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Let $$a_n=\frac{n^22^n}{n!}$$ for all $n\ge1$. Note that each $a_n$ is positive, and that \begin{align}\frac{a_{n+1}}{a_n} &=\cfrac{\left(\frac{(n+1)^22^{n+1}}{(n+1)!}\right)}{\left(\frac{n^22^n}{n!}\right)}\\ &= \frac{(n+1)^2}{n^2}\cdot\frac{2^{n+1}}{2^n}\cdot\frac{n!}{(n+1)!}\\ &= \frac{(n+1)^2}{n^2}\cdot2\cdot\frac{1}{n+1}\\ &= \frac{2n+2}{n^2}\\ &\leq\frac{4n}{n^2}\\ &=\frac4n\end{align} for each $n$. In particular, then, $$a_{n+1}\leq \frac45a_n$$ for all $n\geq 5$, so that $$a_n\leq\left(\frac45\right)^{n-5}a_5$$ for all $n\geq 5$ by simple induction. Now use the Squeeze Theorem.

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This is D'Alambert in disguise. It is good to show how simple it really gets! +1 – Pedro Tamaroff Feb 20 '13 at 20:20

Use the squeeze theorem. You know that it is $\geq 0$ for each $n$. Now, by looking at the independent factors try to find an upper bound that goes to $0$:

$$\frac{n^22^n}{n!}=2^3\frac{n2^{n-3}}{(n-1)!}$$

We have $n-2$ factors both in the numerator and denominator (I omit the $1$ in the factorial)

This gets us, after $n=4$, to

\eqalign{ & {{4 \cdot {2^1}} \over {3!}} = {4 \over 3}{2 \over 2} \le {4 \over 3}{2 \over 2} \cr & {{5 \cdot {2^2}} \over {4!}} = {5 \over 4}{2 \over 3}{2 \over 2} \le {5 \over 4}{2 \over 3} \cr & {{6 \cdot {2^3}} \over {5!}} = {6 \over 5}{2 \over 4}{2 \over 3}{2 \over 2} \le {6 \over 5}{2 \over 4} \cr}

So in general, we can see

$$a_n\leq \frac{16n}{(n-2)(n-1)}$$

for $n\geq 4$. This means that $a_n\to 0$, by the squeeze theorem.

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