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.
 A: In the context of probability theory, you can write
$$
\Pr(\varnothing) = \Pr(\varnothing \cup \varnothing) = \Pr(\varnothing)+\Pr(\varnothing).
$$
Then if you subtract $\Pr(\varnothing)$ from both sides, you get
$$
0 = \Pr(\varnothing).
$$
Thus you don't need infinite series.
That $\infty\cdot0=0$ does not make sense in certain broad contexts, since for example as $x\to\infty$ you have $1/x\to0$ and yet $x\cdot(1/x)\not\to0$; rather $x\cdot(1/x)\to1$ as $x\to\infty$.
However, in many contexts in the theory of measure and integration $0\cdot\infty$ is taken to be $0$.  For example, if you integrate $0$ over a space of infinite measure, you get $0$.
If $c>0$ one can show that $\lim\limits_{n\to\infty} nc\ne c$ by saying that if $\varepsilon=1$, there is no value of $N$ such that for all $n\ge N$ you have $nc$ differing from $c$ by less than $\varepsilon$.  Similarly if $c<0$.
Infinite sums are defined the same way in measure theory as in secondary school:
$$
\sum_{n=1}^\infty a_n = \lim_{N\to\infty}\sum_{n=1}^N a_n,
$$
except that when all $a_n$ are non-negative, one sometimes also uses this definition:
$$
\sum_{n=1}^\infty a_n = \sup\left\{ \sum_{n\in A} a_n : A\subseteq\{1,2,3,\ldots\}\ \&\ A \text{ is finite.} \right\}
$$
But the two definitions do not conflict with each other.
From your comments it appears that you may be trying to prove that finite additivity follows from countable additivity (so that it need not be assumed as a separate axiom). One can write
$$
\Pr(A\cup B) = \Pr(A\cup B\cup\varnothing\cup\varnothing\cup\cdots) = \Pr(A)+\Pr(B)+0+0+0+\cdots
$$
But this works only if you have $\Pr(\varnothing)=0$.  You can show that by proceeding as in the paragraph above that begins with "If $c>0$". 
A: Note that your limit is $\lim_{n\to\infty} n \, c$. If $c \ne 0$, the sequence diverges, so the limit cannot be $c$.
A: You want to prove the statement:
$$\lim_{n\to\infty}\sum_{i=1}^{n}c=c \implies c=0$$

Instead, you can prove the equivalent statement:
$$c\neq0 \implies \lim_{n\to\infty}\sum_{i=1}^{n}c \neq c$$

And this is rather simple, as you can use the exact trick that you were trying to avoid:
$$c\neq0 \implies \lim_{n\to\infty}\sum_{i=1}^{n}c=c\lim_{n\to\infty}\sum_{i=1}^{n}1=c\cdot\infty \neq c$$
A: 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
$$
c = 0
$$
A: $\sum_{i=0}^n c = nc$ 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 = $c$.
