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western50 · Sep 29, 2010, 05:38 PM

1.determine if the sequence converges or diverges, if it converges, find the limit.

a_n=(sin(2n))/(1+square root n)

a_n=n!/(2n)!

a_n=(x^n)/n! (x can be any number)

2. Let a_n be a sequence of positive number such that a_n bigger than a_n+1 for every n. Show that the sequence converges. Give an example of such a sequence that converges to 6.

Unknown008 · Sep 29, 2010, 10:33 PM

1.

$a_n = \frac{sin(2n)}{1+\sqrt{n}}$

The limits of sin(2n) and -1 and 1. What can you say as n gets bigger?

$a_n = \frac{n!}{(2n)!}$

Any idea of how this one might tend to as n gets bigger? Using n = 1 and n = 2 quickly gives a good idea of how the sequence progress.

$a_n = \frac{x^n}{n!}$

There are different outcomes here. If x is negative, there is one pattern.
If x is zero, there is no real pattern.
If x is a fraction, or is equal to 1, there is another pattern
If x is a number greater than 1, there is yet another pattern...

2. I'm not sure about how to prove this one and will as such leave it as be. A simple example though, is:

$a_n = \frac{1}{n} + 6$

As n increases, 1/n tends to zero, and 0+6 = 6.

galactus · Sep 30, 2010, 02:43 AM

For what its worth, if I may give my two cents, here is 1c:

If x=0, the result follows immediately.
Let $a_{n}=\frac{|x|^{n}}{n!}, \;\ n\neq 0$ . Then, $\frac{a_{n+1}}{a_{n}}=\frac{|x|}{n+1}<1 \;\ for \;\ n>|x|$ ,
so $a_{n}$ is eventually decreasing; 0 is a lower bound for $a_{n}$ , so $a_{n}\to L \;\ as \;\ n\to {\infty}$ .
Use $\frac{a_{n}}{a_{n+1}}=\frac{|x|}{n+1}$ to get

$a_{n+1}=\frac{|x|}{n+1}a_{n}$ .

Thus, $L=0\cdot L=0$ .

But, $-|x^{n}|\leq x^{n}\leq |x^{n}|, \;\ so \;\ \frac{|x|^{n}}{n!}\leq \frac{x^{n}}{n!}\leq \frac{|x|^{n}}{n!}$ .

Thus, by the Squeezing Theorem, $\frac{x^{n}}{n!}\to 0 \;\ as \;\ n\to {\infty} \;\ because \;\ \frac{|x|^{n}}{n!}\to 0 \;\ as \;\ n\to {\infty}$

Unknown008 · Sep 30, 2010, 03:10 AM

Originally Posted by galactus

But, $-|x^{n}|\leq x^{n}\leq |x^{n}|, \;\ so \;\ -\frac{|x|^{n}}{n!}\leq \frac{x^{n}}{n!}\leq \frac{|x|^{n}}{n!}$ .

Did you mean this? ^

I think you forgot to put the minus sign in the second inequality.

galactus · Sep 30, 2010, 03:30 AM

No, I do not see a minus sign missing.

Unknown008 · Sep 30, 2010, 03:33 AM

So... when you divided $-|x^{n}|\leq x^{n}\leq |x^{n}|$ by n! the first negative sign is removed? :confused:

galactus · Oct 1, 2010, 04:14 AM

Oh, I see what you mean now, U.

No, that is a typo. I left out the negative by mistake. Sorry, I had to look over it close to see it.

Unknown008 · Oct 1, 2010, 08:53 AM

It's okay :)