Is this space normed or not? This particular question was part of my analysis assignment I was unable to prove / disprove it.

Let $V$ denote the vector space of all sequences $a=(a_1 , a_2,...,a_n,...)$ of real numbers such that $\sum_{n\geq 1} 2^n |a_n|$ converges. Define ||.||: $V \to \mathbb{R}$ by $||a|| =\sum_{n\geq 1}2^n |a_n|$. Then is $V$ complete normed space or not.

There were four questions to be answered regarding $V$ . I have done 3 but unable to do this. I know about completely normed space but I was unable to do this . The  idea is to show that every sequence of sequence converges to a limit and that limit is to be constructed or guessed but I am unable to do it
Kindly guide on how this should be done
 A: Let $l_1=\{(\alpha_1,\alpha_2,...): \sum_{n=1}^{\infty}|\alpha_n|<\infty\}$ and let $T:l_1\to V$ with
$$T(\alpha_1,\alpha_2,...,\alpha_n,...)=\bigl(\frac{1}{2}\alpha_1,\frac{1}{2^2}\alpha_2,...,\frac{1}{2^n}\alpha_n,...\bigr)$$
By the identity
$$||T(\alpha)||_V=\sum_{n=1}^{\infty}2^n|\frac{\alpha_n}{2^n}|=\sum_{n=1}^{\infty}|\alpha_n|=||\alpha||_{l_1}$$
$T$ is a well defined linear and bounded operator from $l_1$ to $V$. Now observe that $T^{-1}(x_1,x_2,...,x_n,...)=(2x_1,2^2x_2,...,2^nx_n,...)$, which means that $T$ is one to one and onto.
Now, for $x=(x_n)_n \in V$ we have
$$||T^{-1}(x)||_{l_1}=\sum_{n=1}^{\infty}2^n|x_n|=||x||_V$$
which means that $T^{-1}$ is also bounded. Hence, $V$ and $l_1$ are isometrically isomorphic. Which means that $V$ is complete.
A: Define a measure space as $(\mathbb N,\mu),$ where for any subset $E\subset \mathbb N,$
$$\mu(E)=\sum_{n\in E} 2^n.$$
Then $L^1(\mathbb N,\mu)$ is precisely the set $V$ you describe, with
$$\|a\|=\int_{\mathbb N} |a|\,d\mu =\|a\|_{L^1(\mathbb N,\mu)}.$$
Since every $L^1$ space is a complete normed linear space, we are done.
Of course, if you haven't studied measure theory, the above is not too useful for this problem. But it's still worth knowing.
A: I'm giving a more basic proof (considering the first two answers). The technique I'm using is rather standard and tries to infer global convergence of a Cauchy family of sequences by analyzing some sort of vertical convergence given componentwise. Let's first prove that  $(V, \| \cdot \|)$ is a normed space. This means that we have to prove first that $\| \cdot \|$ is, in fact, a norm:

*

*Let $a \in V$, if $a \equiv 0$ then $\| a \| = 0$. On the other hand, if $\| a \| = 0$ we can infer that $a \equiv 0$, because otherwise we'll have $a_k \neq 0$ for some $k$ and thus $|a_k| > 0$ which for instance means $\sum_n {2^n} |a_n| \geq {2^k} | a_k| > 0$, a contradiction. The fact that $\| a \| \geq 0$ for all $a \in V$ is trivial. With what we prove, we also have $\| a \| = 0$ iff $a \equiv 0$.

*Let $\beta \in \mathbb{R}$ by the fact that  $| \beta a_k |
 = |\beta | |a_k|$ we have, that for any $a \in V$ $\| \beta a \| = | \beta | \| a \|$.  Where $\beta a = (\beta a_1, ..., \beta a_n, ... )$.

*The triangle inequality follows the same logic. For, let $a, b \in V$ then, since $|a_k + b_k| \leq |a_k| + |b_k|$ (in the real line) then
$\| a + b \| = \sum_k {2^k} | a_k + b_k| \leq \sum_k {2^k}(|a_k| + |b_k|)$
We conclude with the fact that $\sum_k {2^k}|a_k|, \sum_k {2^k}|b_k| < \infty$.

The steps above ensure that $(V, |\ \cdot \|)$ is a normed space. We need to prove now that this space is complete and we are doing it the long way:
Let $\{a^{(n)}\}_n$ be a Cauchy sequence (sequence of sequences) in $(V, \| \cdot \|)$, here we denote the sequence $a^{(k)}$ by $a^{(k)} = (a_{k1}, a_{k2}, ..., a_{km}, ... )$. I see this kind of sequence as a neverending matrix with components given by $a_{ij}$, where we know that its rows tend to be very similar as we go downwards. The trick here is to find whether there is a way to describe the limit of those rows and then see if it lies in $(V, \| \cdot \|)$.
Let $\varepsilon > 0$, then there exists an $N_{\varepsilon} > 0$ such that for any $n,m > N_{\varepsilon}$ we have:
$$\| a^{(n)} - a^{(m)} \| 
< \varepsilon \implies \sum_k {2^k} |a_{nk} - a_{mk}| < \varepsilon$$
Which tell us ${2^k} | a_{nk} - a_{mk} | < \varepsilon$ for any $k \geq 1$. With this in mind, let's define the (vertical) sequences $\{ a_{nk}\}_n$ , and let $\varepsilon_0 > 0$, if we define $\varepsilon = \varepsilon_0 \cdot 2^k$ then, by the above result ($a^{(n)}$ being Cauchy), we have that there exists an $N_{\varepsilon} > 0$ such that for all $n,m > N_{\varepsilon}$ we have
$$| a_{nk} - a_{mk} | <  \varepsilon / 2^k = \varepsilon_0$$
as the above expression is true for any (fixed) $k$, we have that all sequences of the form  $\{ a_{nk}\}_n$ are Cauchy in $(\mathbb{R}, | \cdot |)$ and thus convergent. We can now define the (candidate vertical limit) sequence $b = (b_1, b_2, ..., b_n, ... )$ where $b_k = \lim_{n \to \infty} a_{nk}$. If we manage to prove that $b \in V$ then we finally have that the space is complete.
Let's check if $b \in V$, for that we need $\| b \| < \infty$. In fact,
\begin{equation}
\label{eq:1}
\tag{$\star$}
\|b \| - \| a^{(n)} \| \leq \| b - a^{(n)} \|
\end{equation}
for each $n$, we can zoom in the last term to see
$$\| b - a^{(n)} \| = \sum_k 2^k | b_{k} - a_{nk} |$$
As $\{ a_{nk} \}_n$ converges to $b_k$ for any fixed $k$, we can take $\varepsilon_k = {1} / ({2^k \cdot k^2})$ and find an $N_k$, such that for any $n \geq N_k$, it happens that $2^k | b_{k} - a_{nk} | < 1 / k^2$. With this in mind, let's define $N_l = \sup_{k = 1,..., l} N_k$ and $S_l(n) = \sum_{k = 1}^l 2^k |b_k - a_{nk}|$, then
$$
S_l(N_l) = \sum_{k = 1}^l 2^k |b_k - a_{N_l k}| < \sum_{k=1}^{l} \frac{1}{k^2}
$$
By definition, the sequence $N_l$ is non decreasing in $l$, if it's bounded, then there exists an $N$ such that for any $n \geq N$; $\| b - a^{(n)} \| = \lim_{l \to \infty} S_l(n) <  \sum_{k\geq 1} \frac{1}{2^k} < \infty$. On the other hand, if $N_l$ is not bounded, then as $\lim_{l \to \infty } N_l = \infty$ we substitute $l$ by $N_l$ finding that:
$$
\lim_{N_l \to \infty} \| b - a_{N_l} \| = \lim_{N_l \to \infty} S_{N_l}(N_l) < \sum_{k \geq 1} \frac{1}{k^2} < \infty
$$
Then there exists an $N > 0$ such that,
$$
\| b - a^{(N)} \| < \infty
$$
We conclude using \ref{eq:1}.
