Show the following space is Banach Show that $(X, \| \cdot\|_\star)$ is a Banach space, where $X$ is the following linear space:
$$
X = \left\{ f : \mathbb{N}_{\ge 1} \to \mathbb{R} \mid \|f\|_\star := \left( \sum_{k=1}^\infty (k+1) |f(k)|^2 \right)^{1/2} < \infty \right\}.
$$
Here's my work: Let $\epsilon>0$ be given. Let $(f_n)$ be Cauchy in $X$. Then, there is $N \in \mathbb{N}$ such that whenever $n,m \ge N$, we have
$$
(\forall k) \quad   |f_n(k) - f_m(k)| \le \|f_n - f_m\|_\star < \epsilon
$$
Since $\mathbb{R}$ is complete, each $f_n(k)$ converges to some $a_k \in \mathbb{R}$. Define a new function $f$ by $f(k) = a_k$. Then, we want to show $\| f_n - f\|_\star \to 0$. But this is where I'm stuck.
$$
\|f_n - f\|_\star^2 = \sum_{k=1}^\infty (k+1) |f_n(k) - a_k|^2 \stackrel{?}{\to} 0
$$ 
The rate of convergence for $f_n(k) \to a_k$ obviously depends on $k$, so I can't take a supremum. Moreover, I see no way to push the limit within the sum. 
The final step might depend on the $k+1$, which I haven't made use of.
Or maybe I have approached the problem incorrectly, but this seems quite natural to me . . . 
Thank you for the help.
 A: Continuing where you left off, you have two things to show:


*

*Does the new sequence $f$ belong to $X$?


For this, note that $(f_n)$ is Cauchy and hence norm-bounded. Thus, $\exists R>0$ such that
$$
\sum_{k=0}^{\infty} (k+1)|f_n(k)|^2 \leq R \quad\forall n\in \mathbb{N}
$$
For each $m\in \mathbb{N}$, this implies
$$
\sum_{k=0}^m (k+1)|f_n(k)|^2 \leq R
$$
Let $n\to \infty$ to conclude that
$$
\sum_{k=0}^m (k+1)|f(k)|^2 \leq R 
$$
This is true for all $m \in \mathbb{N}$, so letting $m\to \infty$, you see that $f\in X$ as required.


*Does $\|f_n - f\|_{\ast} \to 0$?


For this, once again use a similar trick: For $\epsilon > 0, \exists N_0 \in \mathbb{N}$ such that
$$
\|f_n - f_l\|_{\ast} < \epsilon \quad\forall n,l\geq N_0
$$
In particular, for $m \in \mathbb{N}$ fixed
$$
\sum_{k=0}^m (k+1)|f_n(k) - f_l(k)|^2 < \epsilon
$$
Let $l \to \infty$, and then $m \to \infty$ to conclude that $\|f_n - f\|_{\ast} \leq \epsilon$ for $n\geq N_0$ as required.

This is all very reminiscent of proving that the $\ell^p$'s are complete, which is why I recommended exploiting that in my remark.
