$\sum_{n=1}^\infty a_n$ is convergent $\iff \lim_{k\to \infty}[\sum_{n=k}^\infty a_n]=0$. Is this right? 
$\displaystyle \sum_{n=1}^\infty a_n $ is convergent $\iff $ $\displaystyle \lim_{k\to \infty}\bigg[\displaystyle \sum_{n=k}^\infty a_n\bigg ]=0$

I don't know whether this statement is true or not. I think I'm sure that the from left to right statement is right, and here is my thought:
$\displaystyle \sum_{n=1}^\infty a_n $ is convergent, and suppose $\displaystyle \sum_{n=1}^\infty a_n=A$. Then $\displaystyle \lim_{k\to \infty}\bigg[\displaystyle \sum_{n=k}^\infty a_n\bigg ]=A-\lim_{p\to \infty}\displaystyle \sum_{n=1}^p a_n=A-A=0$. Thus from left to right is right.
However, I'm not sure about the other side. My intuition tells me the other side might be right. By the same expression, $\displaystyle \lim_{k\to \infty}\bigg[\displaystyle \sum_{n=k}^\infty a_n\bigg ]=\displaystyle \sum_{n=1}^\infty a_n-\lim_{p\to \infty}\displaystyle \sum_{n=1}^p a_n$. If the $\displaystyle \sum_{n=1}^\infty a_n$ isn't convergent. Then the result should not exist instead of being zero, but I'm not sure.
Any help on this? Thanks!
 A: Yes, the theorem is true for real series. As already noted, a series converges iff its partial sums sequence converges (to a finite limit, to be crystal clear). Thus, we put
$$S_n:=\sum_{k=1}^n a_k\;,\;\;\text{so that}\;\;\sum_{k=1}^\infty a_k\;\;\text{converges}\;\iff \lim_{n\to\infty}S_n=S\in\Bbb R$$
and by the very definition of limit we get:
$$\left\{S_n\right\}_{n=1}^\infty\;\;\text{converges to a number}\;\,S\,\;\iff \forall\,\epsilon>0\;\;\exists\,N\in\Bbb N\;\;s.t.\;n>N\implies |S-S_n|<\epsilon$$
But the last inequality is just
$$\epsilon>|S-S_n|=\left|\sum_{k=n+1}^\infty a_k\right|$$
Thus we get that
$$\;\sum_{k=1}^\infty a_k\;\;\text{converges}\;\; \iff \;\lim_{n\to\infty}\sum_{k=n+1}^\infty a_k=0$$
In fact, the above is just the definition of (finite) existence of limit after using
Cauchy's condition for the sequence of partial sums.
A: The main issue with your formula is that, in the right-hand side, we are confronted with an expression $\sum_{n=k}^{\infty} a_n$ that is undefined in some instances.
1. To make an analogy, consider the following formula:
$$ \forall x \in \mathbb{R} \ : \ x > 0 \iff 1/x > 0 \tag{1} $$
For the formula $\text{(1)}$ to have a truth value in formal logic, the function symbol $1/x$ must always have a value, even when $x = 0$. Of course, this is not the case under the usual definition of the division. So, how is this formula dealt with in logic? One solution is to say that $\text{(1)}$ is true iff it is true in any possible interpretation of the symbol $1/x$ (i.e., any assignment of value to the expression $1/0$). However, it is easy to see that the truth value of $\text{(1)}$ depends on the interpretation of $1/0$.
To make amend of this situation, we might tweak $\text{(1)}$ and instead consider
$$ \forall x \in \mathbb{R} \ : \ x > 0 \iff (x \neq 0 \text{ and } 1/x > 0). \tag{1'} $$
This statement is now true in arbitrary interpretation of the symbol $1/0$, hence we can serenely consider it true.
2. In the same spirit, we may modify your formula as:
$$ \sum_{n=1}^{\infty} a_n \text{ converges in } \mathbb{R} \quad \iff \quad \left\{\begin{gathered} \sum_{n=k}^{\infty} a_n \text{ exists in } \overline{\mathbb{R}} \text{ for all } k \text{ and} \\\lim_{k\to\infty} \sum_{n=k}^{\infty} a_n = 0\end{gathered}\right\} $$
where $\overline{\mathbb{R}}=[-\infty,\infty]$. Then it is easy to check that this statement is true, as extensively discussed by other users.
A: Note that convergence of series is usually defined as follows
$$\sum_{k=1}^{\infty}a_k \text{ is convergent } :\Leftrightarrow \left(\sum_{k=1}^n a_k\right)_{n} \text{ is converging}$$
Note that convergence of the term on the right hand side means: Let $\varepsilon>0$ there exists $N\in \mathbb{N}$ such that for all $n\geq N$ it holds $\left|\sum_{k=n+1}^{\infty}a_k\right|\leq \varepsilon$. Then you have
$$ \left|\sum_{k=1}^{\infty} a_k -\sum_{k=1}^n a_k\right|=\left|\sum_{k=n+1}^{\infty}a_k\right|\leq \varepsilon.$$
Since $\varepsilon$ is arbitrary it follows $\sum_{k=1}^n a_k\to \sum_{k=1}^{\infty} a_k$ so that $\sum_{k=1}^{\infty}a_k$ is converging.
