Question

how to show sequence an=a(n-1)+1/2^n is a cauchy sequence?

Answer 1

Given \(\epsilon>0.\) Choose \(N\) so \(\frac{1}{2^N}<\frac{\epsilon}{2}\) . Then, for any \(m>N\) and \(n>N\), we have,
\[\left|a_n-a_{m}\right|=\left| \frac{1}{2^n}-\frac{1}{2^m}\right|< \left| \frac{1}{2^n}\right|+\left|\frac{1}{2^m}\right| < \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon \]
which proves it is a Cauchy sequence.

Answer 2

Sorry, the question was hard to read and I misread it.

Given \(\epsilon>0\) and choose \(N\) so \(\frac{1}{2^N}< \epsilon \) . We can assume \(n\geq m.\) Then, for any \(m>N\) and \(n>N\), we have,
\[\left|a_n-a_{m}\right|= |(a_n-a_{n-1})+(a_{n-1}-a_{n-2})+(a_{n-2}-a_{n-3}) + ...+(a_{m+1}-a_{m})|\]\[\leq |a_n-a_{n-1}|+|a_{n-1}-a_{n-2}|+|a_{n-2}--a_{n-3}| + ...+|a_{m+1}-a_{m}|\]\[ \leq \frac{1}{2^n}+\frac{1}{2^{n-1}}+\frac{1}{2^{n-2}}+ ... +\frac{1}{2^{m+1}}=\frac{1}{2^{m+1}}\left(1+ \frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{n-m-1}}\right)\leq \frac{1}{2^m}\leq \epsilon\]which proves it is a Cauchy sequence.

Answer 3

Question:
Let q(n)=1/2 if n is prime and -1 for othrwise and let {\[a_{n}\]} be defined recursively by \[a_{1}\]=1,\[a_{n+1}\]=\[a_{n}+2^{-n}q(n)\],\[n \ge1\],show that \[a_{n}\] is a cauchy sequences

Answer 4

the previous question is based on this question,but i gave the wrong an

Answer 5

Yes, and the proof is pretty much the same deal.