Vrftsjtryk

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.

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 $(f_n)$.