Sequences... help!

**Rui** · May 18th 2008, 02:15 AM

Consider the sequence {an} = {n!/nⁿ}
a) Write n! = 1*2*...*n-1*n and nⁿ = n*n *....*n*n to express an as a product of n terms.
b) Show that 0≤ an≤ 1/n and use the Squeeze Theorem to show that the sequence converges.
c) Show that the sequence {cⁿ/n!} converges for c a positive constant.
d) Show that the sequence {nⁿ/((2n)!)} converges.

I don't get any of this help!

**Moo** · May 18th 2008, 02:29 AM

Hello,

Originally Posted by Rui

Consider the sequence {an} = {n!/nⁿ}
a) Write n! = 1*2*...*n-1*n and nⁿ = n*n *....*n*n to express an as a product of n terms.

$\frac{n!}{n^n}=\frac{{\color{red}1}*2*{\color{blue }3}*\dots*{\color{magenta}n}}{{\color{red}n}*n*{\c olor{blue}n}\dots*{\color{magenta}n}}= \underbrace{{\color{red}\frac 1n} * \frac 2n * {\color{blue}\frac 3n} * \dots * {\color{magenta}\frac nn}}_{\text{n terms}}$

b) Show that 0≤ an≤ 1/n and use the Squeeze Theorem to show that the sequence converges.

You can easily show it's superior to 0.
Now, you see that the first term is 1/n.
All the terms following 1/n are inferior to 1.

$a_n=\frac 1n * \underbrace{\frac 2n}_{<1} * \underbrace{\frac 3n}_{<1} * \dots * \underbrace{\frac nn}_{\le 1}$

Hence $a_n \le \frac 1n$

And because it's bounded by sequences which converge, this converges.

**Rui** · May 18th 2008, 04:05 AM

Thanks so much Moo, I kinda get it now. But I still don't have any idea on how to do c) and d). Any help would be great.

**Moo** · May 18th 2008, 04:44 AM

Hm... I didn't see it was for College/University, so I was a bit lost

The same way, we write this as a product with n terms.

$\frac{c^n}{n!}=\frac c1 * \frac c2 * \frac c3 * \dots * \frac cn$

Since we work while n tends to infinity, it'll be useful to say that there exists N such that for all n>N, n>c.

$\lim_{n \to \infty} \frac{c^n}{n!}=\underbrace{\frac c1 * \frac c2 * \frac c3 * \dots * \frac c {N-1}}_{\color{red}\text{finite number of terms } \rightarrow \text{ finite value=X}} * \underbrace{\frac cN}_{<1} * \underbrace{\frac c {N+1}}_{<1} * \dots$

So $0 \le \lim_{n \to \infty} \frac{c^n}{n!} \le {\color{red}X}$

--> The sequence converges.

I think it's correct, but I have a very slight hesitation..

**Mathstud28** · May 18th 2008, 06:27 AM

Originally Posted by Rui

Consider the sequence {an} = {n!/nⁿ}
a) Write n! = 1*2*...*n-1*n and nⁿ = n*n *....*n*n to express an as a product of n terms.
b) Show that 0≤ an≤ 1/n and use the Squeeze Theorem to show that the sequence converges.
c) Show that the sequence {cⁿ/n!} converges for c a positive constant.
d) Show that the sequence {nⁿ/((2n)!)} converges.

I don't get any of this help!

For part D I see the most enlightening answer comes from seeing that

$\forall{n}>0,n^n\leq(n!)^2$

and that $\forall{n}>0,\frac{a^n}{n^n}\leq\frac{n^n}{(2n)!}$

So then taking the limits does not change the inequality so coming from the above statement we can see that

$\lim_{n\to\infty}\frac{a^n}{(2n)!}\leq\lim_{n\to\i nfty}\frac{n^n}{(2n)!}\leq\lim_{n\to\infty}\frac{( n!)^2}{(2n)!}$

Now we consider the limits as $n\to\infty$ for each of these functions

The limit of the left side obviously goes to zero considering its an exponential function over an exponential function and $a\ll{n}$ as $n\to\infty$

Alternatively for the rigth side you could Let $L=\lim_{n\to\infty}\frac{a^n}{n^n}=\lim_{n\to\inft y}\bigg(\frac{a}{n}\bigg)^n$

Now taking the natural log of both sides we get

$\ln(L)=\lim_{n\to\infty}\ln\bigg[\bigg(\frac{a}{n}\bigg)^n\bigg]=\lim_{n\to\infty}n\ln\bigg(\frac{a}{n}\bigg)$

Now looking at this we see that direct substitution gives

$\ln(L)=\infty\cdot{-\infty}=-\infty$

NOTE This product yielding -∞ may be too informal for your school. At mine it would be excepted

So we have that $\ln(L)=-\infty\Rightarrow{L=e^{-\infty}=0}$

Thus $\lim_{n\to\infty}\frac{a^n}{n^n}=0$

Now we consider $\lim_{n\to\infty}\frac{(n!)^2}{(2n)!}$

There are many advanced ways to do this but the most intutive and easiest to show way is to do this

Expanding we get

$\lim_{n\to\infty}\frac{\bigg(1\cdot{2}\cdot{3}\cdo t...\cdot{(n-1)}\cdot{(n)}\bigg)\bigg(1\cdot{2}\cdot{3}\cdot... \cdot{(n-1)}\cdot{(n)}\bigg)}{1\cdot{2}\cdot{3}\cdot...\cdo t{(n-1)}\cdot{n}\cdot...\cdot{(2n-1)}\cdot{(2n)}}$

Cancelling we get

$\lim_{n\to\infty}\frac{1\cdot{2}\cdot{3}\cdot....\ cdot{(n-1)}\cdot{(n)}}{(n+1)\cdot{(n+2)}\cdot{(n+3)}\cdot. ..\cdot{(2n-1)}\cdot{(2n)}}$

Now it is clear to see that the denominator as $n\to\infty$ is infinitely greater than the numerator thus making

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

So now we look back to our original problem

$\lim_{n\to\infty}\frac{a^n}{n^n}\leq\lim_{n\to\inf ty}\frac{n^n}{(2n)!}\leq\lim_{n\to\infty}\frac{(n! )^2}{(2n)!}$

So then seeing as we showed that $\lim_{n\to\infty}\frac{a^n}{n^n}=0$

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

We can rewrite this inequality to read

$0\leq\lim_{n\to\infty}\frac{n^n}{(2n)!}\leq{0}$

Then by the squeeze theorem we conclude that

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

And therefore is convergent

**Moo** · May 18th 2008, 06:35 AM

Originally Posted by Mathstud28

For part D I see the most enlightening answer comes from seeing that

$\forall{n}>0,n^n\leq(n!)^2$

and that $\forall{n}>0,\frac{a^n}{n^n}\leq\frac{n^n}{(2n)!}$

And where does it come from ?

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