# $(E,\|.\|)$ is complete $\iff$ every normally convergent series is convergent.

$(E,\|.\|)$ is complete $\iff$ every normally convergent series is convergent.

A series $\sum_{n=1}^{\infty} u_n$ is normally convergent $\iff$ $\sum_{n=1}^{\infty} \|u_n\|$ converges.

$(\Rightarrow)$

Suppose $(E,\|.\|)$ complete and $\sum_{n=1}^{\infty} \|u_n\|$ convergent. Then letting $\epsilon > 0$, $\exists N$ such that $\sum_{n=N}^{\infty} \|u_n\| < \epsilon$ (1)

The series $\sum_{n=1}^{\infty} u_n$ will converge if the sequence $(s_n) = (u_1+\cdots+u_n)$ converge.

$\|s_n - s_m\| = \|\sum_{n=1}^{\infty} u_n- \sum_{m=1}^{\infty} u_m\| = \|\sum_{n=m+1}^{n} u_n\| \leq \sum_{n=m+1}^{n} \|u_n\|$

(1) implies $\sum_{n=m+1}^{n} \|u_n\| < \epsilon$. Then $(s_n)$ is Cauchy, since $E$ is complete, $(s_n)$ converges.

$(\Leftarrow)$

Suppose $\sum_{n=1}^{\infty} \|u_n\|$ convergent $\Rightarrow \sum_{n=1}^{\infty} u_n$ convergent.

Suppose $(x_n)$ is a Cauchy sequence in $E$. If there exists a subsequence $(x_{n_k})$ convergent, then $(x_n)$ is convergent.

Since $(x_n)$ is Cauchy, $(x_{n_k})$ is Cauchy, and there exists a $n_0 \in \mathbb{N}$ such that $n_k, n_{k+1}>n_0$ implies $\|x_{n_{k+1}}-x_{n_k}\| < 2^{-k}$ . Define the following sequence: $(y_n) = (x_{n_1},x_{n_2}-x_{n_1},\ldots,x_{n_k}-x_{n_{k-1}},\ldots)$

Then: $\sum_{n=1}^{\infty} \| y_n\| = \|x_{n_1} + x_{n_2}-x_{n_1} + \cdots + x_{n_k} - x_{n_{k-1}} + \cdots \| \leq \|x_{n_1}\| + \|x_{n_2}-x_{n_1}\| + \cdots + \|x_{n_k} - x_{n_{k-1}}\| + \cdots = \|x_{n_1}\| + \sum_{k=1}^{\infty} 2^{-k} = \|x_{n-1}\|+1$. Hence, by hypothesis, $\sum_{n=1}^{\infty} y_n$ is convergent. Then the sequence $(y_1,y_1+y_2,y_1+y_2+y_3,\ldots) = (x_{n_1},x_{n_2},x_{n_3},\ldots)$ is convergent. $\square$.

Please verify if my attempt is correct and I also would like to see an alternate proof for the converse $(\Leftarrow)$ part. Thanks.

Looks fine but perhaps in the last part start with a finite sum $$\sum_{n=1}^N \|y_n\| = ... \leq \|x_{n_1}\| + 1$$ and then take the $N\rightarrow \infty$ limit. It's a fairly standard proof and I don't think there are good alternatives.