# How can I show that $\sum_{n=1}^\infty x_n$ converges in a Banach space?

Let $$X$$ be a Banach space and $$(x_n)_n\subset X$$ a sequence s.t. $$\sum_{n\ge 1}||x_n||<\infty$$. I want to show that $$\sum_{n=1}^\infty x_n$$ converges in $$X$$.

Proof Since $$\sum_{n\ge 1}||x_n||<\infty$$, we deduce that $$||x_n||\rightarrow 0$$, thus for all $$\epsilon >0$$ there exists $$N\in \Bbb{N}$$ such that for all $$n\geq N$$, $$||x_n||<\frac{\epsilon}{l-(k+1)}$$for some $$l,k\geq N$$.

Now consider $$S_m=\sum_{n=1}^m x_n$$. Let $$\epsilon >0$$ and pick $$M=N$$, then for all $$l,k\geq M$$ (assuming W.L.O.G. that $$k)$$||S_l-S_k||=\left|\left|\sum_{n=k+1}^l x_n\right|\right|\leq\sum_{n=k+1}^l ||x_n||\leq \sum_{n=k+1}^l \frac{\epsilon}{l-(k+1)}=\epsilon$$So in particular $$(S_m)_m$$ is a cauchy sequence and since $$X$$ is complete we conclude that $$\lim_{m\rightarrow \infty}S_m\sum_{n=1}^\infty x_n$$ converges in $$X$$.

Is this argument valid?

• No, that's not a valid argument. The bound you chose for $\|x_n\|$ is true for some $k,l \geq N$, not for all. Jan 16 at 20:53
• @GiorgosGiapitzakis why isn't it for all $k,l$? Jan 16 at 20:54
• Because that's how you defined it to be. Also, the way your proof is structured, you're only using the (weaker) fact that $\|x_n\|\to 0$, which of course is not enough to prove that the initial series converges. Jan 16 at 20:56
• @GiorgosGiapitzakis and how would one do it then? Jan 16 at 20:58

You need to use the stronger fact that $$\sum_{n=1}^\infty \|x_n\| < +\infty$$ whereas in your proof you're only using the weaker implication that $$\|x_n\|\to 0$$. With that assumption alone (i.e replacing the convergence of $$\sum_n \|x_n\|$$ to just $$\|x_n\|\to 0$$), the result doesn't hold. An easy counterexample can be realized by taking $$X=(\mathbb{R},|\cdot |)$$ and $$x_n = 1/n.$$
Now to prove the claim, you'll need to use the fact that the convergence of the series of norms implies that the sequence $$T_n = \|x_1\| + \dots + \|x_n\|$$ is Cauchy. So for $$\varepsilon > 0$$ there exists some $$N\in \mathbb{N}$$ such that for all $$n, m \geq N$$ $$\sum_{i=n+1}^m \|x_i\| < \varepsilon$$ The triangle inequality shows that $$S_n = x_1 + \dots + x_n$$ is Cauchy and therefore converges (due to $$X$$ being a Banach space).
• but $\sum_{n=1}^\infty 1/n$ does not converge Jan 16 at 21:07
• @user123234 That's why the assumption $\|x_n\| \to 0$ alone is not enough (which is what you were using in your proof). Jan 16 at 21:09
• but then even the assumption $\sum ||x_n||<\infty$ is not satisfied so we cannot say anything Jan 16 at 21:09
• Sorry if I wasn't clear. What I'm trying to say is that by weakening the assumption from the convergence of series of norms to just $\|x_n\|\to 0$ the result doesn't still hold. Jan 16 at 21:10