How to show that space is complete? Let $N_\alpha=\{(x_n)_{n=1}^\infty\mid \sum_{j=1}^n |x_j|\leq Mn^\alpha\}$, where $\alpha\in R$. Show that $N_\alpha$ is Banach space with the norm $\|(x_n)_{n=1}^\infty\|=\sup_{n\in N} n^{-\alpha} \sum_{j=1}^n|x_j|$. 
It is is easy to check properties of norm, but I don't know how to prove completeness of space $N_\alpha$ (or, that every Cauchy sequence converges). Any kind of help is welcome.
 A: Suppose $x_n$ is Cauchy. We need to first produce a candidate limit, and then show that this limit is in $N_\alpha$.
(I am assuming that you have shown the various properties of the norm, in particular, $\|x+y\| \le \|x\|+ \|y\|$.)
Then for all $\epsilon>0$ there exists $N$ such that if $n,m \ge N$ then, for all $p$, $\sum_{k=1}^p |[x_n]_k-[x_m]_k| \le \epsilon p^\alpha$. In particular, for a fixed $p$, $|[x_n]_p-[x_m]_p| \le \epsilon p^\alpha$. Hence $[x_n]_k$ is Cauchy and converges to a limit. Let $[\hat{x}]_k = \lim_{n \to \infty} [x_n]_k$. The sequence $\hat{x}$ is our candidate limit.
Now we must show that $\hat{x} \in N_{\alpha}$.
Now fix $\epsilon =1$, and choose $N$ such that if $m,n \ge N$, then $\sum_{k=1}^p |[x_n]_k-[x_m]_k| \le  p^\alpha$. If we fix $p$ and then take the limit as $n \to \infty$, and set $m=N$, we have $\sum_{k=1}^p |[\hat{x}]_k-[x_N]_k| \le  p^\alpha$.
If we let $[y]_k = [\hat{x}]_k-[x_N]_k$, the above shows that $y \in N_\alpha$.
Since $\|x_N + y\| \le \|x_N\|+ \|y\|$, we have $x_N + y \in N_\alpha$, and since $[\hat{x}]_k = [x_N]_k + [y]_k$, we see that $\|\hat{x} \| \le \|x_N\|+ \|y\| < \infty$, hence $\hat{x} \in N_\alpha$.
