Proofs of Boundedness

[daon]

I have a bunch of these to do, but here are two:

1) \(\displaystyle \L x_n = \sum_{i=0}^n \frac{1}{2^i}\) is increasing and bounded above by 2.
I think I have shown that it is increasing correctly, by showing that x_n+1 > x_n. But how do I show it is bounded above by two? I figured that I can show that since $\displaystyle \sum_{i=0}^n (\frac{1}{2^i}) = 1 + \sum_{i=1}^n (\frac{1}{2^i})$, so it would be sufficient to show that $\displaystyle \sum_{i=1}^n (\frac{1}{2^i})$ is bounded by 1. However I am not sure how to do this...

2) \(\displaystyle \L \sum ^n_{i=0} \frac{1}{i!}\). I think I will be able to get this after I understand how to do the first one.

Thanks

 

[JakeD]

For (1) you can use the formula for a geometric series $\displaystyle \large \sum_{i=0}^n r^i = \frac{1-r^{n+1}}{1-r}$ with $\displaystyle r = 1/2.$ This is always less than 2 for that value of $\displaystyle r.$

I have no help for (2).

 

[pka]

Daon, for #2 take note of the following and use the basic comparison test.
\(\displaystyle \L
n > 3\quad \Rightarrow \quad 2^n < n!\quad \Rightarrow \quad \frac{1}{{n!}} < \frac{1}{{2^n }}\)

 

[daon]

I don't have these tests available to me. I believe I need to prove them with something like the basic definition of a limit (not the delta one), though we are not limited to this. I feel proving that 2 is the limit would prove that it is an upper bound, or not?

So, say I let $\displaystyle \epsilon > 0$ then there exists an N $\displaystyle \in \mathbb{N}$ s.t. for all n > N, \(\displaystyle \L |\sum ^n _{i=0} \frac{1}{2^i} - 2| < \epsilon\)

And, sorry, for number 2 I had left out some info. I need to show that it is bounded above by 3. i.e. bounded by the first sum + 1. That is how I intend to show it. (edit: pka, I just now realized that is what you said!)