# I'm not sure I'm correct: $\|\sum_{n=1}^\infty x_n\| \leq \sum_{n=1}^\infty \|x_n\|$

I was solving this Functional Analysis problem, but I'm not sure I'm correct on this one, the problem is:

Let $\big(E,\|\bullet\|\big)$ be a normed vector space, and $(x_n)_n$ a sequence in $E$ such that $\sum_{n=1}^\infty x_n$ converges. Show that $$\left\|\sum_{n=1}^\infty x_n\right\| \leq \sum_{n=1}^\infty\|x_n\|$$

My solution:

We know that the following inequality holds for any $m\in\mathbb{N}$ (just using the triangular inequality):

$$\left\|\sum_{n=1}^m x_n\right\| \leq \sum_{n=1}^m \|x_n\|$$

Since the left side converges, the inequality still holds in the limit $m\rightarrow\infty$, hence we have the result.

My doubts:

The only thing I am not totally sure is my last statement, because I can't know if the right side converges, so how can I compare the two things?

But then I thought I can use the following argument:

The sequence $\left(\sum_{n=1}^m \|x_n\|\right)_m$ is clearly a monotone (non-decreasing) real sequence, so if it is bounded it converges and my result is correct, the other option is that the sequence is unbounded, in which case

$$\sum_{n=1}^\infty \|x_n\|=\infty$$

And then it makes sense to write:

$$\left\|\sum_{n=1}^\infty x_n\right\| < \infty$$

Can someone tell me if my reasoning is correct? Or if another way would be better to solve this problem?

• Thank you for a question that is well written, and you also give your own thoughts. +1 – Eff Sep 9 '17 at 13:00

If $\sum_{n=1}^\infty \|x_n\|=\infty$ then it is true. So let us assume that $\sum_{n=1}^\infty \|x_n\|<\infty$. So, $$\left\|\sum_{n=1}^m x_n\right\| \leq \sum_{n=1}^m \|x_n\| \le \left\|\sum_{n=1}^m x_n\right\| \leq \sum_{n=1}^m \|x_n\|=l$$ Since, norm is a continuous function so $$\lim_{m\to \infty}\left\|\sum_{n=1}^m x_n\right\|=\left\|\lim_{m\to\infty} \sum_{n=1}^m x_n\right\| =\left\|\sum_{n=1}^\infty x_n\right\| \le \lim_{m\to \infty} l=l$$
You're quite right that the right hand side need not converge : a simple example in the reals is $x_n = \frac{(-1)^n}{n}$ where the series converges, but not absolutely. So you can indeed make the distinction between $+\infty$ and finite for the right hand series. $+\infty$ is no problem for the inequality, and the finite case follows from the triangle inequality, as said. Or just use (in your first version) that in the extended reals $x_n \le y_n$, $x\to x, y\to y$ also implies $x \le y$ (this holds in all ordered topological spaces, as $\{(p,q) \in X^2: p \le q \}$ is closed in $X^2$). The right hand limit always exists by monotonicity (and completeness of $[-\infty, +\infty]$), he left hand side as $\sim_n x_n$ converges and $\|.\|$ is continuous.
Using the continuity of $\|\cdot\|$, we have: $$\left\|\sum_{n=1}^\infty x_n\right\| = \left\|\lim_{m\to\infty}\sum_{n=1}^mx_n\right\| = \lim_{m\to\infty}\left\|\sum_{n=1}^m x_n\right\| \leq \lim_{m\to\infty}\sum_{n=1}^m \left\|x_n\right\| = \sum_{n=1}^\infty \left\|x_n\right\|$$