Approximating the limit of a Cauchy sequence in a Banach space

Question

Let $E$ be a Banach space and consider a sequence $(x_n)_n$ in $E$ satisfying the following condition: $$||x_n-x_{n-1}||\leq 3^{-n}\mbox{ for all }n\in\mathbb{N}.$$ Clearly $(x_n)_n$ is a Cauchy sequence and therefore converges to an element $x\in E$.

Question: How to prove that $||x-x_n||\leq \frac{1}{2}3^{-n}$ for all $n\in\mathbb{N}$?

Answer 1

Let $m > n$. We have:

\begin{align} \|x_m - x_n\| &\le \|x_{m} - x_{m-1}\| + \cdots + \|x_{n+1} - x_n\| \\ &\le \frac1{3^m} + \cdots + \frac1{3^{n+1}}\\ &= \frac1{3^{n+1}}\left(1 + \frac13 + \cdots + \frac1{3^{m-n-1}}\right)\\ &= \frac1{3^{n+1}} \frac{1 - \frac1{3^{m-n}}}{1 - \frac13} \end{align}

Letting $m\to\infty$ we get

$$\|x - x_n\| \le \frac1{3^{n+1}} \frac1{1 - \frac13} = \frac12 \cdot 3^{-n}$$

Answer 2

This is basically just the triangle inequality: the distance from $x_n$ to $x$ is at most the distance we traverse by going through all the terms of the sequence approaching $x$. For $N\geq n$, we have $$\|x_N-x_n\|\leq \|x_N-x_{N-1}\|+\|x_{N-1}-x_{N-2}\|+\dots+\|x_{n+1}-x_n\|\leq 3^{-N}+3^{-N+1}+\dots+3^{-n-1}.$$ Taking the limit as $N\to\infty$, we get $$\|x-x_n\|\leq \sum_{k=n+1}^\infty 3^{-k}=\frac{1}{2}3^{-n}.$$

Answer 3

We can prove it using triangle equation: $$||x-x_n|| \leq \sum_{k=n}^{\infty}||x_{k+1}-x_k|| \leq \sum_{k=n}^{\infty}3^{-k-1}=\frac{3^{-n-1}}{1-\frac{1}{3}} = \frac{3^{-n}}{2}$$