how do I find the upper and lower bound of this sequence? (2^(n-1))/(n+1)!

Well, the ratio test should tell you that this sequence is null; i.e. the limit is zero. Since all of the terms are positive, you should be able to get a lower bound trivially.

 

Moreover, [imath]a_{n+1} - a_n = \frac{2^{n-1}}{(n+1)!} \left( \frac{2}{n+2} - 1\right) < 0[/imath] for any n > 0. So the sequence is strictly decreasing, and it should be easy to find an upper bound from here.

I got (2)/(n!(n+2)) so as n increases to infinity, the thing approaches 0 so that is why it is decreasing right? and less than 1 means it converges.

but I don't understand where the upper/lower bound, what is it.... suppose to be a finite number? or is it like lower bound 0 since it is approaching 0 and upper bound unbounded?

Your sequence is a series of discrete points [math](a_1, a_2, a_3, \dots)[/math]. An upper bound is a number M such that [math]a_n \leq M[/math] for any N. A lower bound is a number m such that [math]a_n \geq m[/math] for any N. Notice that upper and lower bounds have no dependency on n at all.

 

Now, yes, your sequence does converge to zero. But that doesn't mean that it's decreasing. For example, take [imath]a_n = \frac{(-1)^n}{n}[/imath]. This is a sequence which tends to zero, but is not strictly decreasing. To show a sequence is stricly decreasing, you need to show that [imath]a_{n+1} - a_n < 0[/imath] for all n - which is what I have done above.

 

Basically, now that you know it's strictly decreasing, you know that you're not going to have places where the sequence might jump up and then start to fall down again, so you can say that the upper bound for the sequence is simply [imath]a_1[/imath]. The lower bound will trivially be zero.

Incidentally I've been talking about "the" bounds for the sequence. In fact, for a bounded sequence there are infinitely many upper and lower bounds. When I say "the" upper bound, I'm actually talking about the least upper bound - i.e. the supremum, if you've covered that at all.

so basically, if there is an upper and lower bound for this type of sequence, it is either a1 and where it converges?

Basically, yes. More formally, if [imath]a_n[/imath] is a decreasing sequence, then [imath]a_1 \geq a_n[/imath] for all n, and if, moreover, it is convergent then [imath]\lim_{n\to\infty} a_n \leq a_n[/imath].