Consider the vector space V of all real sequences such that $\sum_{n=1}^{\infty}2^n|a_n|$ converges. Consider the norm as $||(a_n)||= \sum_{n=1}^{\infty}2^n|a_n|$
- Is this vector space complete?
- Is the dimension countable?
For part 1. I considered the sequence of sequences with finitely many positions = $(1/2)^n$ from 1st to nth position and rest 0. I "believe" under this norm this sequence converges to the sequence with nth element = $(1/2)^n$ and then the sum will not converge. Am I right?
For part 2. I am stumped. But I do see that it is at least not finite dimensional.
The sequence that you created does not converge in that norm.
The answer has been presented by @Mindlack in the comment. But I would like to point out a way to see the completeness of $V$.
Actually $V$ is $L^{1}(\mu)$, where $d\mu(x)=\omega(x)d\nu(x)$, here the variable $x$ is discrete, it is just $n$, and $\omega(x)=2^{x}$, $\nu$ being the counting measure, and that \begin{align*} \sum_{n}2^{n}|a_{n}|=\int_{\mathbb{N}}|f(x)|\omega(x)d\nu(x)=\int_{\mathbb{N}}|f(x)|d\mu(x), \end{align*} where $f(x)=a_{x}$. And we know that $L^{1}(\mu)$ is complete for every measure $\mu$.