Is it true that if $\lim_{x\rightarrow ∞}x_{k}=0$ then $\sum_{k=1}^{∞}x_{k}$ is convergent? 
Let $E$ be a normed vector space and $\sum_\limits{k=1}^{\infty}x_{k}$ a series in $E$.
  Is it true that if $\lim_\limits{x\to\infty}x_{k}=0$ then $\sum_\limits{k=1}^{\infty}x_{k}$ is convergent?

My attempt:
It is obviously not true. Let $\sum_\limits{k=1}^{\infty}\frac{1}{k}$, we know that $\lim\limits_{x\to\infty}\frac{1}{k}=0$ but $\sum_\limits{k=1}^{\infty}\frac{1}{k}$ is not convergent.
Does anyone know if there's a more formal way to prove this?
 A: The great thing about proving that something is false is that you "only" have to provide a single counter-example or prove that a counter-example have to exist. Formally this follows (as already pointed out by egreg in the comments) from taking the negation of the statement $[\forall x\,P(x)]$ (for all $x$ the statement $P(x)$ is true) which becomes $[\exists x\,\neg P(x)]$ (there exist an $x$ such that the statement $P(x)$ is false). For this particular problem the negation of the problem statement can be stated as follows: 

"There exists a normal vector space $E$ and a sequence $a_n\in E$ with $\lim_{n\to\infty} a_n = 0$ where $\sum a_n$ diverges"

What you have done is therefore perfectly fine: $E = \mathbb{R}$ with the absolute-value norm $|\cdot|$ is an example of a normed vector-space and as you correctly say the series $\sum \frac{1}{n}$ diverges even though $\lim_{n\to\infty}\frac{1}{n} = 0$. The only thing you might choose to add to make the proof more self-contained is a proof that your counter-example is a real counter-example, i.e. that $\sum \frac{1}{n}$ diverges. However in this case one might argue that this series is so well known that it's not really needed here (the proof can be found in practically all calculus textbooks and in hundreds of questions on this site).
