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 $k<l$)$$||S_l-S_k||=\left|\left|\sum_{n=k+1}^l x_n\right|\right|\leq\sum_{n=k+1}^l ||x_n||\leq \sum_{n=k+1}^l \frac{\epsilon}{l-(k+1)}=\epsilon$$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?

  • $\begingroup$ 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. $\endgroup$ Jan 16 at 20:53
  • $\begingroup$ @GiorgosGiapitzakis why isn't it for all $k,l$? $\endgroup$
    – user123234
    Jan 16 at 20:54
  • $\begingroup$ 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. $\endgroup$ Jan 16 at 20:56
  • $\begingroup$ @GiorgosGiapitzakis and how would one do it then? $\endgroup$
    – user123234
    Jan 16 at 20:58

1 Answer 1


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).

  • $\begingroup$ but $\sum_{n=1}^\infty 1/n$ does not converge $\endgroup$
    – user123234
    Jan 16 at 21:07
  • $\begingroup$ @user123234 That's why the assumption $\|x_n\| \to 0$ alone is not enough (which is what you were using in your proof). $\endgroup$ Jan 16 at 21:09
  • $\begingroup$ but then even the assumption $\sum ||x_n||<\infty$ is not satisfied so we cannot say anything $\endgroup$
    – user123234
    Jan 16 at 21:09
  • $\begingroup$ 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. $\endgroup$ Jan 16 at 21:10
  • 1
    $\begingroup$ ah okey, thanks a lot! $\endgroup$
    – user123234
    Jan 16 at 21:11

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