# Convergence in sequence space

Let the sequence space $s = \{$ all sequences of complex numbers $\}$ with distance $$d(x,y) = \sum_{j=1}^{\infty} \frac{1}{2^j} \frac{ | \xi_j - \eta_j| }{ 1 + |\xi_j - \eta_j|}.$$ Let $x_n = (\xi_j^{(n)})=(\xi_1^{(n)}, \xi_2^{(n)},...)$, $x= (\xi_1, \xi_2, ...)$.

I want to show that $x_n \to x$ iff $\xi_j^{(n)} \to \xi_j \quad \forall j$

For the first implication, suppose $x_n \to x$. Fix $\epsilon > 0.$ There exists $N$ s.t. $n\geq N$ implies $d(x_n, x) < \epsilon$, i.e. $$d(x_n,x) = \sum_{j=1}^{\infty} \frac{1}{2^j} \frac{ | \xi_j^{(n)} - \xi_j| }{ 1 + |\xi_j^{(n)} - \xi_j|} < \epsilon$$

Fix $j$ . I want to show that $| \xi_j^{(n)} - \xi_j| < \epsilon$. I know that : $$\frac{1}{2^j} \frac{ | \xi_j^{(n)} - \xi_j| }{ 1 + |\xi_j^{(n)} - \xi_j|} < \epsilon$$

What should I do next? Thanks!

## 1 Answer

For proving the convergence $\xi_j^{(n)}\to\xi_j$, let's better use the definition of $x_n\to x$ with $\frac{1}{2^{j+2}}>\frac{\epsilon}{2^{j+1}}>0$ then, as you did

$$\frac{1}{2^j}\frac{|\xi_j^{(n)}-\xi_j|}{1+|\xi_j^{(n)}-\xi_j|}<\frac{\epsilon}{2^{j+1}}$$

from where $\frac{|\xi_j^{(n)}-\xi_j|}{1+|\xi_j^{(n)}-\xi_j|}<\frac{\epsilon}{2}$, i.e.

$$|\xi_j^{(n)}-\xi_j|<\frac{\epsilon/2}{1-\epsilon/2}\leq\epsilon$$

Now we need the other direction. Assume each $\xi_j^{(n)}\to\xi_j$ and let $\epsilon>0$.

Observe that the quotients $\frac{|X|}{1+|X|}<1$. Therefore there is $J\in\mathbb{N}$ such that $\sum_{j>J}\frac{1}{2^j}\frac{|\xi_j^{(n)}-\xi_j|}{1+|\xi_j^{(n)-\xi_j}|}<\frac{\epsilon}{2}$.

Now choose $N\in\mathbb{N}$ such that $|\xi_j^{(n)}-\xi_j|<\min\{\frac{\epsilon}{2},1\}$ for $0<j\leq J$ and $n>N$.

then \begin{align}d(x_n,x)&=\sum_{i\leq j}\frac{1}{2^j}\frac{|\xi_j^{(n)}-\xi_j|}{1+|\xi_j^{(n)}-\xi_j|}+\sum_{j>J}\frac{1}{2^j}\frac{|\xi_j^{(n)}-\xi_j|}{1+|\xi_j^{(n)}-\xi_j|}\\&\leq\sum_{i\leq j}\frac{1}{2^{j+1}}|\xi_j^{(n)}-\xi_j|+\sum_{j>J}\frac{1}{2^j}\frac{|\xi_j^{(n)}-\xi_j|}{1+|\xi_j^{(n)}-\xi_j|}\\&\leq\frac{\epsilon}{2}\sum_{j\leq J}\frac{1}{2^{j+1}}+\frac{\epsilon}{2}\\&\leq\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\end{align}

• the inequality $\frac{\epsilon}{1-\epsilon} \leq \epsilon$ is false if $\epsilon < 1$, in which case it would be equivalent to $\epsilon \leq \epsilon - \epsilon ^2$ – user159517 Jan 21 '15 at 20:34