# Show that the sequence $(\frac{2^n}{n!})$ has a limit.

Show that the sequence $(\frac{2^n}{n!})$ has a limit.

I initially inferred that the question required me to use the definition of the limit of a sequence because a sequence is convergent if it has a limit $$\left|\frac{2^n}{n!} - L \right|{}{}<{}{}\epsilon$$

I've come across approaches that use the squeeze theorem but I'm not sure whether its applicable to my question. While I have found answers on this site to similar questions containing the sequence, they all assume the limit is 0.

I think I need to show $a_n ≥ a_{n+1},\forall n \geq 1$, so

$$a_{n+1} = \frac{2^{n+1}}{(n+1)!}=\frac{2}{n+1}\frac{2^{n}}{n!}<a_n$$

A monotonic decreasing sequence is convergent and this particular sequence is bounded below by zero since its terms are postive. I'm not sure whether or not I need to do more to answer the question.

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What is your question? –  Phira Dec 5 '12 at 10:01

It is easy to prove that $$\sum_{k=1}^{\infty}\frac{2^n}{n!} \lt \infty$$ e.g. with the ratio test you have $$\frac{a_{n+1}}{a_n}=\frac{n!}{2^n}\cdot\frac{2^{n+1}}{(n+1)!}=\frac{2}{n+1}\longrightarrow 0$$ Then $\lim\limits_{n\rightarrow\infty}\frac{2^n}{n!}$ has to be $0$

If you do not know anything about series you might assert, that $n!>3^n$ if $n\gt N_0$ for some fixed $N_0\in\mathbb{N}$. Therefore you have for those $n$ $$0<\frac{2^n}{n!} \lt\frac{2^n}{3^n}=\left(\frac{2}{3}\right)^n\longrightarrow 0$$ Thus $$\lim\limits_{n\longrightarrow\infty} \frac{2^n}{n!} =0$$

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I don't see how this is easier than the initial question. Could you develop ? Furthermore, I'm not even sure that user51462 know what a series is. –  Siméon Dec 5 '12 at 10:11
I edited my answer. Now you see a possible solution. In my opinion this is easier than calculating the limit of the given sequence directly. –  stg Dec 5 '12 at 10:18
@Julien B Could you kindly elaborate as to why I appear to not know what a series is? –  user51462 Dec 5 '12 at 10:23
@Fant I agree but the ratio test should also be proven... –  Siméon Dec 5 '12 at 10:26
@JulienB. I agree to your point of view, thus I added a optional solution to my answer –  stg Dec 5 '12 at 12:02

For $n>2$, we have \begin{eqnarray*} 0 < \frac{2^n}{n!}&=&\frac{2}{1}\times\frac{2}{2}\times\cdots\times\frac{2}{n-1}\times\frac{2}{n}\\ &\leq&2\times1\times\cdots\times1\times\frac{2}{n}=\frac{4}{n}.\end{eqnarray*} Then by the squeeze principle (or directly from the definition of limit) we get $$\lim_{n\to\infty}\frac{2^n}{n!}=0.$$

So if you've proven the squeeze principle, you can apply it. If not, you can go directly to the definition:

Let $\epsilon>0$. Then taking $N>\frac{4}{\epsilon}$, we find that $$\left|\frac{2^n}{n!}-0\right| = \frac{2^n}{n!} \leq \frac{4}{n} < \epsilon,\qquad \hbox{for all}\quad n>N.$$ This proves directly that the limit is zero. (Just to tidy up one tiny loose thread: we need to take $N$ bigger than both 2 (so that the inequality worked out above applies) and $4/\epsilon$.)

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Note that each term of the sequence $a_n=\dfrac{2^n}{n!}$ is greater than $0$.

Furthermore, $a_{n+1}=\dfrac2{n+1}a_n$ so that for $n>1$ we have $a_{n+1}\lt a_n$.

By the Monotone Convergence Theorem, a decreasing sequence of reals, bounded below, has a limit.

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The sequence is decreasing and bounded below, hence it has a finite limit say $a_n\to x$. Now $$a_{n+1}=\frac{2}{n+1} a_n$$ Taking limit $n\to\infty$ on both side you have $x=0.x=0$.

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Since $$e^x = \sum_{r=0}^{\infty} \frac{x^r}{r!}$$ for all $x \in \mathbb{R}$, by Cauchy criterion $$\lim_{n \to \infty} \frac{x^n}{n!} = 0$$
$\frac{a_{n+1}}{a_{n}}= \frac{2}{n+1}$ After dividing the fraction by n you get 0 with $n\rightarrow \infty$ and 0 < 1. So the series converges. In my opinion this is enough and definately simplier than Epsilon/Delta criterion :)