If $(a_n)_{n\in\mathbb N}\subseteq[0,\infty)$, what do we need to conclude $\sum\limits_{k=n+1}^\infty a_k\to0$ when $n\to\infty$? Let $(a_n)_{n\in\mathbb N}\subseteq[0,\infty)$. Since $\left(\sum_{k=1}^na_k\right)_{k\in\mathbb N}$ is increasing, $$a:=\sum_{k=1}^\infty a_k=\lim_{n\to\infty}\sum_{k=1}^na_k=\sup_{n\in\mathbb N}\sum_{k=1}^na_k\in[0,\infty]\;.$$

I want to conclude $$\sum_{k=n+1}^\infty a_k\xrightarrow{n\to\infty}0\;.\tag1$$

We should have \begin{equation}\begin{split}\lim_{n\to\infty}\sum_{k=n+1}^\infty a_k&=\lim_{n\to\infty}\lim_{N\to\infty}\sum_{k=n+1}^Na_k\\&=\lim_{n\to\infty}\lim_{N\to\infty}\left(\sum_{k=1}^Na_k-\sum_{k=1}^na_k\right)\\&=\lim_{n\to\infty}\left(a-\sum_{k=1}^na_k\right)=a-a=0\;.\end{split}\tag2\end{equation} I want to stress that I don't think that there is any problem even when $a=\infty$ (since $\infty-\infty=0$). However, in any proof in which I saw things like $(1)$, the author concludes $(1)$ by noting that $a<\infty$. So, what am I missing?
 A: Cauchy's convergence test is a iff class theorem. As a result ($a_n\geq0, \forall n$, this is given):


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*If $\sum\limits_{k=1}^{\infty} a_k < \infty \Rightarrow \forall \varepsilon>0, 
 \exists N \in \mathbb{N}: 0<\left|\sum\limits_{k=n+1}^{n+p} a_k\right|=\sum\limits_{k=n+1}^{n+p} a_k<\varepsilon$ for $\forall n>N$ and $\forall p\geq 1$. Then $$\lim\limits_{p\rightarrow \infty} \sum\limits_{k=n+1}^{n+p} a_k \leq \varepsilon$$
In fact we can start with $\frac{\varepsilon}{2}$ and conclude $$\lim\limits_{p\rightarrow \infty} \sum\limits_{k=n+1}^{n+p} a_k \leq \frac{\varepsilon}{2}<\varepsilon$$
so we conclude that $\forall \varepsilon>0, 
 \exists N \in \mathbb{N}: 0<\sum\limits_{k=n+1}^{\infty} a_k<\varepsilon$ for $\forall n>N$, which means $$\sum\limits_{k=1}^{\infty} a_k < \infty \Rightarrow \lim\limits_{n\rightarrow \infty}\sum\limits_{k=n+1}^{\infty} a_k=0$$

*$\lim\limits_{n\rightarrow \infty}\sum\limits_{k=n+1}^{\infty} a_k=0 \Rightarrow \forall \varepsilon>0, 
 \exists N \in \mathbb{N}: 0<\sum\limits_{k=n+1}^{\infty} a_k<\varepsilon$ for $\forall n>N$. But $$0<\sum\limits_{k=n+1}^{n+p} a_k <\sum\limits_{k=n+1}^{\infty} a_k< \varepsilon$$
for $\forall n>N$ and $\forall p\geq 1$, which means $$\lim\limits_{n\rightarrow \infty}\sum\limits_{k=n+1}^{\infty} a_k=0 \Rightarrow \sum\limits_{k=1}^{\infty} a_k < \infty$$
Altogether 
$$\lim\limits_{n\rightarrow \infty}\sum\limits_{k=n+1}^{\infty} a_k=0 \Leftrightarrow \sum\limits_{k=1}^{\infty} a_k < \infty$$
that's the answer to what we need to conclude ...
