I see! so I make it like this:
\frac{1}{4^N} < \epsilon
\frac{1}{4}\ln\left(\frac{1}{\epsilon}\right) < N
Or in other words, to answer you question I choose:
N>0.25\ln(100)>1.15, so N \geq 2
Thank you so much, sorry for not being any good at the simple part. I'm a mech engineering...
So is it just something like this?
\frac{4}{3}\left(\frac{1}{4^m} - \frac{1}{4^n}\right) \leq \frac{1}{3\cdot4^N}
or even just this:
\frac{4}{3}\left(\frac{1}{4^m} - \frac{1}{4^n}\right) \leq \frac{1}{4^N}
I'm fairly certain we need to make this stuff pretty water tight in this course, so I'm...
Homework Statement
My problem is this, you are given that for a sequence the following is true:
\left|x_{n} - x_{n+1}\right| < \frac{1}{4^{n}}
So its an obvious case to prove the sequence is Cauchy.
Homework Equations
So I'll state the stuff just for the sake of it...