Prove there are infinitely many converging subsequences which do not overlap with each other. Let $||$ a$_n$ -a$_{2n}$$||$ $\leq$ $\frac{1}{2^n}$   for all n$\geq$ 1. Prove there are infinitely many converging subsequences which do not overlap with each other. a$_n$ $\in$ R$^m$
Well, I know a cauchy sequence always conveges. there exists an N such that $||a_n - a_m|| \leq \epsilon $  for all n, m > $N$ 
Still confused how to show say the subsequence {a$_{2^n}$} is Cauchy.
 A: It may help to note that we are doing two rather different things: 
First, we argue that there is a convergent subsequence (the one indexed by powers of two, for which we simply note that $\sum_n \frac1{2^n}$ converges so that $a_1,a_2,a_4,a_8,a_{16},\dots$ is indeed Cauchy). 
Second, we argue that $\mathbb N$ (and therefore any infinite set of indices, such as the set of powers of two) can be split into infinitely many disjoint infinite sets (for which we can use powers of primes, or whatever), each giving rise to an appropriate sub-subsequence (that of course converges to the same limit as the original subsequence of powers of two). 
The point is that these two steps have nothing to do with one another but, combined, give the required result.
A: Hint:
Consider the subsequences index by the powers of prime numbers: $\{a_{p^n}\}_{n \in \mathbb{N}}$. This are trivially non-overlapping and you can try to prove that they converge being Cauhy's subsequences.
$||a_{p^m} - a_{p^n}|| \le ||a_{p^m} - a_{2p^m}|| + ||a_{2p^m} - a_{4p^m}|| + \dots$
EDIT: Let $\epsilon > 0$ be given. $||a_{p^m} - a_{p^n}|| \le ||a_{p^m} - a_{2p^m}|| + ||a_{2p^m} - a_{4p^m}|| + \dots + ||a_{2^kp^m} - a_{p^n}||$.
Here I just used the triangular inequality applied multiple times to $||a_{p^m} - a_{p^n}|| = ||a_{p^m} - a_{2p^m} + a_{2p^m} - a_{4p^m} + \dots -a_{2^kp^m} +a_{2^kp^m} - a_{p^n}||$.
Here $k$ is taken such that $2^kp^m \le p^n$ and $p^n \le 2^{k+1}p^m$.
Now, if you apply the hypothesis $||a_n - a_{2n}|| \le \frac{1}{2^n}$ you have:
$||a_{p^m} - a_{p^n}|| \le \frac{1}{2^{p^m}} + \frac{1}{2^{p^m}} \dots + \frac{1}{2^{p^m}} = \frac{k}{2^{p^m}}$.
To prove that this is a Cauchy sequence we need to prove that $||a_{p^m} - a_{p^n}|| \le \epsilon$, but follow easily if we take $m$ such that $\frac{k}{2^{p^m}} \le \epsilon$.
I hope this helps!
EDIT:
There should be a way to fix things, the easier I actually have is to forget everything and take the following sequences: $\{a_{2^np}\}_{n \in \mathbb{N}}$ where $p$ is a prime number. This time the previous argument applies.
