If $(x_n)$ is a Cauchy sequence, then it has a subsequence such that $\|x_{n_k} - x_{n_{k-1}}\| < 1/2^k$

Question

If $X$ is a normed linear space and $(x_n)$ is a Cauchy sequence in $X$, then $x_n$ has a subsequence $x_{n_k}$ which satisfies $$\|x_{n_k} - x_{n_{k-1}}\| < \frac 1 {2^k}$$ for every $k > 0$.

Why is this assertion true?

Answer 1

The definition of Cauchy sequence begins by saying "for every $\varepsilon>0$". Whatever is true of EVERY positive number is true of $1/2^kn$.

Answer 2

Suppose $(x_n)_{n\in N}$ is a Cauchy sequence. Let $(a_n)_{n\in N}$ be any monotonically decreasing sequence of positive numbers. Let $g(1)$ be the least (or any) $n$ such that $\forall n'>n\;(|x_n-x_{n'}|<a_1).$ Recursively, let $g(j+1)$ be the least (or any) $n>g(j)$ such that $\forall n'>n\;(|x_n-x_{n'}|<a_{n+1}).$ The subsequence $(a_{g(n)})_{n\in N}$ satisfies $\forall n\;(|x_{g(n)}-x_{g(n+1)}|<a_n)$. In particular we may choose $a_n=2^{n+1}$ for each $n\in N$.