Monotone sequences update

[JJ007]

1.) Show that the given sequence is eventually strictly increasing or decreasing.
\(\displaystyle \frac {\ n!}{3^n}\limits_{n=1}^\infty\)
$\displaystyle \frac {\ a_{n+1}}{a_n} = \frac {(\ n+1)!}{3^{n+1}} * \frac {\ 3^n}{n!} =?$



2.) Show: $\displaystyle \frac {\ 1}{1*3}+ \frac {\ 1}{2*4}+ \frac {\ 1}{3*5}+...= \frac {\ 3}{4}$

Telescopic? $\displaystyle S_n = \sum\limits_{k=1}^n \frac {\ 1}{(2k-1)(2k+1)}= \frac {\ 1}{2} \sum\limits_{k=1}^n \frac {\ 1}{(2k-1)} - \sum\limits_{k=1}^n \frac {\ 1}{2k+1}=?$

Thanks

 

[galactus]

Re: Monotone sequences Infinite Series

1.) Show that the given sequence is eventually strictly increasing or decreasing.
\(\displaystyle \frac {\ n!}{3^n}\limits_{n=1}^\infty\)
$\displaystyle \frac {\ a_{n+1}}{a_n} = \frac {(\ n+1)!}{3^{n+1}} * \frac {\ 3^n}{n!} =?$

$\displaystyle \lim_{n\to \infty}\frac{(n+1)!}{3^{n+1}}\cdot\frac{3^{n}}{n!}=\lim_{n\to \infty}\frac{n+1}{3}={\infty}$

2.) Show: $\displaystyle \frac {\ 1}{1*3}+ \frac {\ 1}{2*4}+ \frac {\ 1}{3*5}+...= \frac {\ 3}{4}$

Telescopic? $\displaystyle S_n = \sum\limits_{k=1}^n \frac {\ 1}{(2k-1)(2k+1)}= \frac {\ 1}{2} \sum\limits_{k=1}^n \frac {\ 1}{(2k-1)} - \sum\limits_{k=1}^n \frac {\ 1}{2k+1}=?$

Thanks

Expand to $\displaystyle \sum_{k=1}^{\infty}\frac{1}{k(k+2)}=\sum_{k=1}^{\infty}\left[\frac{1}{2k}-\frac{1}{2(k+2)}\right]$

This gives the telescopic series you seek.

$\displaystyle \left(\frac{1}{2}-\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+\left(\frac{1}{6}-\frac{1}{10}\right)+\left(\frac{1}{8}-\frac{1}{12}\right)+.......$

As can be seen, all of the terms cancel one another except the 1/2+1/4=3/4. And that's the sum.

 

[JJ007]

Re: Monotone sequences Infinite Series

$\displaystyle \lim_{n\to \infty}\frac{(n+1)!}{3^{n+1}}\cdot\frac{3^{n}}{n!}=\lim_{n\to \infty}\frac{n+1}{3}={\infty}$

Ok, so what canceled out?

Expand to $\displaystyle \sum_{k=1}^{\infty}\frac{1}{k(k+2)}=\sum_{k=1}^{\infty}\left[\frac{1}{2k}-\frac{1}{2(k+2)}\right]$

You multiplied by 1/2 right?

 

[JJ007]

Re: Monotone sequences Infinite Series

Also my second solution was based on this problem:
Show $\displaystyle \frac {\ 1}{1*3}+ \frac {\ 1}{3*5}+ \frac {\ 1}{5*7}+...=\frac {\ 1}{2}$

My apologies. Although your solution was helpful for that problem is it any different for this one?

 

[Deleted member 4993]

Re: Monotone sequences Infinite Series

Also my second solution was based on this problem:

Show $\displaystyle \frac {\ 1}{1*3}+ \frac {\ 1}{3*5}+ \frac {\ 1}{5*7}+...=\frac {\ 1}{2}$ ? same principle - why don't you try it and tell us

My apologies. Although your solution was helpful for that problem is it any different for this one?

 

[Deleted member 4993]

Re: Monotone sequences Infinite Series

$\displaystyle \lim_{n\to \infty}\frac{(n+1)!}{3^{n+1}}\cdot\frac{3^{n}}{n!}=\lim_{n\to \infty}\frac{n+1}{3}={\infty}$

Ok, so what canceled out?

(n+1)! = (n+1) * n!

3[sup:38759wna]n+1[/sup:38759wna] = 3 * 3[sup:38759wna]n[/sup:38759wna]

 

[JJ007]

Re: Monotone sequences Infinite Series

New modified problem.

$\displaystyle a_n= \frac {\ 5^n}{n!}$
Show that the sequence $\displaystyle a_n$ is bounded

$\displaystyle \frac {\ 5^n}{n!}=\frac {\ 5^{n\times n\dots\times n}}{n\times(n-1)\times(n-2)\times\dots\times 2\times 1} ?$

It follows that the sequence $\displaystyle {a_n}$ is convergent. Why? Determine its limit.

$\displaystyle \frac{a_{n+1}}{a_n}= \frac{\ 5}{n+1}?$

Thanks