Here is my try..

By hypothesis, the convergence of $\sum_{i=1}^{\infty}\|x_n\|$ in $\mathbb{R}$ $\implies$ the convergence of $\sum_{i=1}^{\infty} x_n$ in $X$.

Pick a Cauchy sequence $\{ f_n\}$ in $X$. Then $\forall$ $\epsilon > 0$, $\exists$ $n>m>N$ s.t $\|f_n - f_m \|< \epsilon$.

and $| \|f_n\| - \|f_m\| | \leq \|f_n - f_m \|< \epsilon$ that is $\{\|f_n\|\}$ is a Cauchy sequence in $\mathbb{R}$ thus it converges in $\mathbb{R}$. Hence it is bounded (i.e. $\exists k>0$ s.t | \|$f_n$\| | $\leq k$, $\forall$ $n=1,2,..$).

$\implies S_n=\sum_{i=1}^{n}\|f_n\| \leq nk$ but also the sequence $\{S_n\}$ is monotonically increasing thus $\{S_n\}$ converges in $\mathbb{R}$ $\implies \sum_{i=1}^{\infty} f_n$ converges in $X$ by hypothesis thus the sequence $f_n \rightarrow 0$ in $X$. Thus $X$ is a Banach space.

  • $\begingroup$ Take a look at the definition of boundedness and then maybe reconsider your assertion that $S_n$ is bounded. ;) $\endgroup$ – MaoWao Nov 16 '18 at 23:46
  • $\begingroup$ Oh yes, $S_n$ can't be bounded as the set of natural numbers is not bounded above. Thank you! $\endgroup$ – Dreamer123 Nov 17 '18 at 0:05
  • $\begingroup$ If you type \|f\| instead of ||f|| it looks like $\|f\|$ instead of $||f||$. $\endgroup$ – DanielWainfleet Nov 17 '18 at 13:57

If a subsequence of a Cauchy sequenec converges, so does the entire sequence. Choose subsequence $f_{n_k}$ such that $\|f_{n_k}-f_{n_{k+1}}\| < \frac 1 {2^{k}}$. Then $\sum_k (f_{n_k}-f_{n_{k+1}})$ converges absolutely and hence it converges. By writing down the partial sums show that $\lim_k f_{n_k}$ exists. This proves convergence of $\{f_{n_k}\}$ hence that of $(f_n)$.

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