Is it possible to represent every irrational number as a (limit of) an infinite sum of rational numbers? For instance, we can certainly represent π in this fashion.

$$
\frac{\pi}{4} \;=\; \sum_{n=0}^\infty \, \frac{(-1)^n}{2n+1} .\!
$$

$\ln(2)$ is also irrational. And even that can be represented as an infinite sum of a sequence of rational numbers:

$$
\ln (1+x) \;=\; \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} x^n.
$$
with $x=1$.

And also, $\sqrt2$:

$$
\sqrt2 \;=\; \sum_{k=0}^\infty\frac{(2k-1)!!}{4^kk!}\tag{2}
$$

I'm curious if this applies to all irrational numbers? Is so, how do you go about proving it?
Since that series
$$
\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}
$$
converges conditionally, the Riemann Rearrangement Theorem says that we can get every real number, rational or irrational, by rearranging the terms of that series

So, yes, every irrational number can be written as the limit of the sum of rational numbers.
The rational numbers are dense in the real numbers.  

Therefore every irrational number x greater than 0, there exists a rational number y such that 

$$0 &lt; y &lt; x $$

and 

$$y + \frac{x-y}{2} + \frac{\frac{x-y}{2}}{2} + \frac{\frac{\frac{x-y}{2}}{2}}{2} ...$$

converges to x.  

(This can easily be generalized for x &lt; 0.)

QED