# If every absolutely convergent series is convergent in $X$, then $X$ is a Banach space.

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.

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