The sum of infinitely many $c$s is $c$ implies $c = 0$. This seems like an obvious claim, but I would like to be able to prove this rigorously.
Suppose I have $c \in \mathbb{R}$ satisfying
$$\lim_{n \to \infty}\sum_{i=1}^{n}c = c\text{.}$$
How does it follow that $c = 0$?
What I think I shouldn't do: $$c\lim_{n \to \infty}n = c$$
because this gives $0 \cdot \infty = 0$ (is this true?).
To understand the context of this question, I am working with a probability measure $\mathbb{P}$ and am trying to show that $\mathbb{P}\left(\emptyset\right) = 0$ using the definition (and in this case, $\mathbb{P}\left(\emptyset\right) = c$). Maybe infinite sums are defined differently in measure theory? I don't know.

We'll assume that $c \in \mathbb{R}$.
$$\lim_{n\to\infty}\sum_{i=1}^{n}c=c$$
Subtract $2c$
$$\lim_{n\to\infty}\sum_{i=1}^{n-2}c=-c$$
$$-\lim_{n\to\infty}\sum_{i=1}^{n-2}c=\lim_{n\to\infty}\sum_{i=1}^{n}c$$
But we know that $$\lim_{n\to\infty}\sum_{i=1}^{n-2}c = \lim_{n\to\infty}\sum_{i=1}^{n}c$$
Thus
$$
-\lim_{n\to\infty}\sum_{i=1}^{n}c = \lim_{n\to\infty}\sum_{i=1}^{n}c
$$
and so we conclude that 
$$
\lim_{n\to\infty}\sum_{i=1}^{n}c = 0
$$
and since we have that
$$
\lim_{n\to\infty}\sum_{i=1}^{n}c=c
$$
then we finally conclude that
$$
c = 0
$$
