Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I know this question was asked before, but I couldn't find it and I think my question is slightly different. I'm new here so if I do something wrong, I'm sorry, I'm still trying to learn.

I need to "study the limit behavior of the sequence and when it exists compute their limits."

This is what I have tried so far:

I studied the first few terms of the sequence, and I claimed that the sequence was monotone and decreasing. Then I showed that $a_{n+1}-a_n\leq0$ by: $$a_{n+1}-a_n= \frac {2^{n+1}} {(n+1)!}\ - \frac {2^n} {n!}= \frac {2^{n+1}} {n!(n+1)}- \frac {2^n} {n!}=\frac {2^n} {n!}(\frac {2} {n+1})-\frac {2^n} {n!}=\frac {2^n} {n!}(\frac {2} {n+1}-1)=\frac {2^n} {n!}(\frac {n-1} {n+1}) $$ Then I said: $\frac {2^n} {n!}(\frac {n-1} {n+1})\leq 0$ only if $\frac {n-1} {n+1}\leq 0$ because $2^n>0$ and $n!>0$ . Then I said that $n+1>0$ for n existing the naturals and $1-n\leq0$, thus $\frac {n−1} {n+1}\leq 0$ and $a_{n+1}-a_n\leq 0$.

Therefore the sequence is monotonic and decreasing. Because the first term is 2 and $\frac {2^n} {n!}>0$, the sequence is bounded and together the monotonicity and and boundedness means that a limit exists.

Now I feel that what I have done so far is correct, and instinctually I know the limit is $0$, but how do I compute that? Do I just claim it is $0$ and prove it using the epsilon definition of a limit?

We haven't covered anything with series, but we have covered cauchy sequences (is that supposed to help me?) and things about lim (inf) and lim (sup). Any help please?

share|improve this question
    
@JavaMan Just realized that I do know about the ratio test. But now I have a more general question, should I have appraoched this problem with the ratio test first instead of checking for monotonicity and boundedness? Also, is there anything specific that alerts me to use the ratio test instead of checking monotonicity/boundedness? –  user66807 Mar 15 '13 at 2:28

7 Answers 7

up vote 7 down vote accepted

I'd suggest taking the $\lim_{n\to \infty}$ of the ratio of two consecutive terms (use the ratio test) : $$\lim_{n\to \infty} \dfrac{a_{n +1}}{a_n}\; = \lim_{n\to \infty}\dfrac{\dfrac{2^{n+1}}{(n+1)!}}{\dfrac{2^n}{n!}} $$ $$= \lim_{n\to \infty} \dfrac{2^{n+1}n!}{2^n(n+1)!} = \lim_{n\to \infty} \dfrac{2}{n+1}$$

Evaluate the limit: If the limit exists, and $L \lt 1$, then as DonAntonio aptly clarified, "the series whose general term is the given sequence converges," by the Ratio Test for series. And then by Convergence test for series, if its series converges, the limit of the sequence itself (the general term of the series), is equal to $0$.


share|improve this answer
1  
Yes, that is what I meant, Thanks @DonAntonio: I will edit my post and credit your phrasing! –  amWhy Mar 15 '13 at 2:59
    
I came here but you were gone. + –  Babak S. Mar 15 '13 at 5:28

For every $x \in \mathbb{C}$ it is well-known that $\sum\limits_{n=0}^{\infty} \frac{x^n}{n!}$ converges to $\mathrm{exp}(x)$, hence $\frac{x^n}{n!}$ converges to $0$.

share|improve this answer

Recall: For a sequence $\{u_n\}$ in $\mathbb R^+,$

  • $\lim \dfrac{u_{n+1}}{u_n}<1\implies \lim u_n=0;$

  • $\lim \dfrac{u_{n+1}}{u_n}> 1\implies \lim u_n=+\infty;$

  • For $\lim \dfrac{u_{n+1}}{u_n}=1$ no definite conclusion can be made e.g. consider $\left\{\dfrac{n+1}{n}\right\}$ and $\left\{\dfrac{1}{n}\right\}.$

Here, $\dfrac{a_{n+1}}{a_n}=\dfrac{2}{n+1}\to0<1.$ Therefore $\{a_n\}$ converges to $0.$

share|improve this answer

A few ways to look at it.

The sum of all $a_n$ is finite (consider the power series for $e^x$ at $x=2$).

For $n>5$ every term is at most half the previous one.

$2^n$ grows quickly, but not nearly as quickly as $n!$.

share|improve this answer

$$\frac{2^n}{n!}=\frac{2}{1}[\frac{2}{2}\frac{2}{3}..\frac{2}{n-1}]\frac{2}{n} \leq \frac{2}{1}\cdot[ 1 \cdot 1..\cdot 1]\frac{2}{n}\leq \frac{4}{n}$$

Since $0 \leq \frac{2^n}{n!} \leq \frac{4}{n}$ the sequence converges to $0$.

share|improve this answer

Six solutions have already been written but I have a different one:

Use the bound* $$ \left(\frac n e\right)^n \le n! \le n^n $$

This shows that $$0 \le \frac {2^n} {n!} \le \left( \frac {2e} n \right)^n \longrightarrow 0$$

giving the result.

*Proof of bound: The upper bound (which I did not use) follows trivially from the definition of the factorial. The lower bound follows from a quick and dirty evaluation of the Gamma function integral,

$$n! = \Gamma(n+1) = \int_0^\infty e^{-t} t^n dt \geq \int_n^\infty e^{-t} t^n dt \geq \int_n^\infty e^{-t} n^n dt = \left( \frac n e \right)^n$$

This is a useful bound of the factorial to know - it's weaker than the Stirling approximation, but much, much easier to prove.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.