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.

$\sum_{i=0}^n c = nc$ so you want to prove $\lim_{n \rightarrow \infty} cn \rightarrow c \iff c = 0$.
If $c \ne 0$ and {$nc$} converges to $d$, then for any $ \epsilon > 0$ I can find an $N_{\epsilon} \in \mathbb N$ such that for any $n > N_{\epsilon} \implies |nc - d| < \epsilon$ .  
So for $0 < \epsilon < |c|/2$ and any $m > n > N_{\epsilon}$, then $|c| \le (m-n)|c| = |mc - nc| \le |mc -d| + |d - nc| < 2\epsilon$.  Which is a contradiction as we chose $\epsilon < |c|/2$.
So {$nc$} can only converge if $c = 0$ and if $c = 0$ then obviously {$nc$} = {0} converges to 0 = $c$.
