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.

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.

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}

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.

No, I do not see a minus sign missing.

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

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.

It's okay :)