Sequence problem

[jessebu]

Consider the sequence {bn} defined by

{bn} = ?(from i=0 to n) 1/(2^i)

Prove it converges by showing it is a Cauchy sequence.

So I have to prove that for all epsilon there is a number n such that if m, n ? N then abs (am - an) < epsilon.
I'm not sure how to get the proof started.

 

[daon]

I'll do the sequence, not the series, and see if you can use my steps:

\(\displaystyle \exists N \in \mathbb{N} \,\, s.t. \,\, 2^{-N} < \epsilon\)

Let \(\displaystyle m \ge n \ge N\)

Then

\(\displaystyle |\frac{1}{2^n}-\frac{1}{2^m}| = 2^{-n}|1-\frac{1}{2^{m-n}}| \le 2^{-n} \le 2^{-N} < \epsilon\)

Try something similar and report back. Here's a hint:

If \(\displaystyle m \ge n\) then

\(\displaystyle |b_m-b_n|=\sum_{i=0}^m \frac{1}{2^i}-\sum_{i=0}^n \frac{1}{2^i} = \sum_{i=n+1}^m \frac{1}{2^i} = (\frac{1}{2^{n+1}} +\frac{1}{2^{n+2}}+ \dots +\frac{1}{2^m})\)

 

[jessebu]

So we can say that (1/(2^(n+1)) + ... + 1/(2^m) < 2^(-n) < 2^(-N) < epsilon.

Correct?
Thanks a lot.

 

[daon]

Well, you need to show there can be such an N. It looks like you basically copied and pasted my conclusion. You should spend some time thinking about it. Why should there be such an N?

Questions like this are not as straight-forward as today's standard Calculus problems. This was the kind of question I'd see in my undergraduate Analysis class. Hence more thought should be given to such a problem.