$(X,||.||)$ a Normed space, prove that when we got the property $\sum x_n$ converges in $X$ if $\sum||x_n||$ converges in $\mathbb{R}$, that X Banach. $(X,||.||)$ a Normed space, prove that when we got the property $\sum x_n$ converges in $X$ if $\sum||x_n||$ converges in $\mathbb{R}$, that X is a Banach Space.
So far i have this:
Let $\{x_n\}$ be a cauchy sequence, then we can construct a sequence such that for $n,m>N_k$:
$||x_n - x_m|| < \frac{1}{2^k}$.
Choose a subsequence $y_k = \{x_s\}_{s \geq N_k}$.
Then
$\sum ||y_k|| \leq \sum \frac{1}{2^k} + ||x_m|| \leq \infty$ 
So we know that $\sum y_k$ is convergent in X.
I know that if a sub sequence of a Cauchy sequence is convergent, the Cauchy sequence is convergent as well.
But how does one now conclude that $y_k$ is a convergent sequence if its sum is convergent...?
 A: Hint: Find a subsequence $x_{n_k}$ such that $\|x_{n_{k+1}} - x_{n_k}\| < 1/2^k, k \in \mathbb N.$ Then recall the standard way of turning a sequence into a series.
A: Your notation is a bit messy but the idea is correct. Construct a subsequence $\{ x_{n_k}\}$ so that $$\lvert \lvert x_{n_{k+1}} - x_{n_k}\rvert \rvert < \frac 1 {2^k}.$$ Then define $y_K$ = $x_{n_0} + \sum^K_{k=0} (x_{n_{k+1}} - x_{n_k})$. You'll find that $\{y_K\}$ is absolutely summable and hence $$\lim_{K\to \infty} \left[ x_{n_0}  + \sum^K_{k=1} (x_{n_{k+1}} - x_{n_k}) \right] = x \text{ for some } x\in X \,\,\,\, \text{(i.e. the limit converges)}.$$ But what does the expression in the brackets simplify to?
A: with User8128 and Zhw's tip:
construct $y_k = \{x_{n_{o}}, x_{n_{1}} - x_{n_{0}},x_{n_{2}} - x_{n_{1}}, \dots  ,x_{n_{k}} - x_{n_{k-1}}\} $
Then we see that: $\sum_{k=0}^\infty ||y_k|| = \sum_{k=0}^\infty ||y_k + x_{n_{k}} - x_{n_{k}}|| \leq \sum_{k=0}^\infty \frac{1}{2^k} + ||x_{n_{k}}|| < \infty$.
So $\sum_{k=0}^\infty y_k$ converges in $X$.
Note now that the partial sums converge as well. So we always get:
$\sum_{k=0}^M y_k = \{x_{n_{m}}\}$
and since this subsequence of a cauchy sequence converges, we are finished $\quad$
