We define for $l^\infty$ (bounded sequences) the following norm (the $n$-th term of the sequence $x$ is $x(n)$): $$ ||x||=\sum_{n=1}^{\infty}\frac{|x(n)|}{2^n} $$ The question is if $(l^\infty, ||\cdot||)$ is a Banach space. A well known result is that a normed space $X$ is a Banach space iff every absolutely convergent series is convergent, so I defined the following sequence in $l^\infty$:
Let $(x_m)$ be the sequence such that $x_m(n)=0$ if $n\neq m$, $x_m(n)=m$ if $n=m$ $$ x_1=(1,0,0,...) \\ x_2=(0,2,0,0,...) \\ x_3=(0,0,3,0,0,...) $$ And so on. Clearly the sequence $x_m$ is bounded (by $m$) for all $m$, so $(x_m)$ is a sequence in $l^\infty$. I claim that the series with general term $x_m$ is absolutely convergent but not convergent, so $(l^\infty, ||\cdot||)$ can't be a Banach space. $$ \sum_{m=1}^{\infty}||x_m||=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{|x_m(n)|}{2^n}=\sum_{m=1}^{\infty}\frac{m}{2^m}=2<\infty $$so there is absolute convergence. However, $$ \sum_{m=1}^{\infty}x_m=(n)=(1,2,3,4,...) $$ which is not bounded. Is this solution correct? Thank you
That is not exactly correct, because you can't write $\sum\limits_{m = 1}^\infty x_m = (1, 2, 3, \ldots)$ without defining in what sense it converges. You need to prove that $\sum\limits_{m = 1}^\infty x_m$ diverges differently.
You can consider extension of your space (sequences such that $\sum\limits_{m = 1}^\infty x_m 2^{-m}$ is bounded), in which this series converges to element not from your space.
Alternatively, you can note that convergence in your space implies coordinate-wise convergence, but $\sum\limits_{m = 1}^N x_m$ can't converge coordinate-wise to any bounded sequence.