1
$\begingroup$

Is the sequence $\left\{\frac{2^n}{n!}\right\}$ convergent? If so, what is the limit?

$$ \frac{2^n}{n!} - 0 = \frac{2^n}{n!} < \frac {2^n}{n} <\; ? < \epsilon$$

I dont know how to simplify $\frac{2^n }{ n}$.

I cannot just do $\frac{2^n}{n}<\epsilon$ right ?

  • 1
    At each step you multiply by $2$ bu divide by $n$, so ... – 2017-02-16

6 Answers 6

10

Consider $\sum \frac{2^n}{n!}$ and do the ratio test:

$$\lim_{n\to\infty} \frac{2^{n+1}}{(n+1)!}\frac{n!}{2^n}=\lim_{n\to\infty}\frac{2}{n+1}=0$$

The ratio test implies that the sum is convergent which implies that the limit is $0$

  • 0
    Nice proof. Nevertheless, note that the proof of the ratio test for series uses d'Alembert criteria which states that if $(x_n)$ is a sequence such that $|x_n/x_{n+1}|\to l<1$, then $x_n\to 0$. So there is in fact no need to pass through series for that (which is a bit overkill I think). Weirdly, the "ratio test for sequence" = "d'Alembert criteria" only [appears in french on wikipedia](https://fr.wikipedia.org/wiki/R%C3%A8gle_de_d'Alembert). – 2017-02-17
5

$\frac{2^n}{n!} = \frac{2}{1} \cdot \frac{2}{2} \cdot\frac{2}{3} \cdot\frac{2}{4} \cdot\frac{2}{5} \dots \frac{2}{n} < \frac{2}{1} \cdot \frac{2}{2} \cdot\frac{2}{3} \cdot\frac{2}{4} \cdot\frac{2}{4} \dots \frac{2}{4} = \frac{8}{6}\cdot(\frac{1}{2})^{n-3}$

4

For $\; n \geq 3 \;$ we have

$$\frac{2^n}{n!} \;\; = \;\; \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot \frac{2}{5} \cdot \frac{2}{6} \cdot \;\; \cdots \;\; \cdot \frac{2}{n} \;\; \leq \;\; \frac{2}{1} \cdot \frac{2}{2} \cdot \left(\frac{2}{3}\right)^{n-2} \;\; \longrightarrow \;\; 0$$

  • 0
    Nice, short argument. +1 – 2017-02-16
3

HINT:

$$n!\ge (n/2)^{n/2}$$

so that

$$\frac{2^n}{n!}\le \left(\frac{8}{n}\right)^{n/2}$$

Alternatively, note that

$$\frac{2^n}{n!}= \left(\frac{2}{1}\right)\,\overbrace{\left(\frac{2}{2}\right)\left(\frac{2}{3}\right)\left(\frac{2}{4}\right)\cdots \left(\frac{2}{n-1}\right)}^{n-2\,\,\text{terms all less than or equal to }1}\,\frac{2}{n}\le \frac4n$$

  • 0
    @DonAntonio Thank you my friend! ;-)) – 2017-02-16
1

the series $\sum_\limits{n=0}^{\infty} \frac {x^n}{n!}$ converges for all x. In fact it equals $e^x$

That means that the sequence $\{\frac {x^n}{n!}\}$ must converge to 0.

  • 0
    I dont know..... but I have to use inequlity to prove it. – 2017-02-16
0

$$ \frac21<\frac22<\frac23<\frac13<\frac2{15}<...<\frac1n$$ So the above sequence is bounded by $1/n$; but $\{\frac1n\} \to 0 \text{ as } n\to\infty$.