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?
 A: Here's a simple answer.
Pick an irrational x.
Find a rational in (x - 1, x). Call this $a_0$.  Find another rational in (x - 1/2, x).  Call this $a_1$.  Keep finding rational $a_n$ in $(x - 1/2^n , x)$.  These $a_n$ converge to x. 
$x=a_0+\sum\limits_{n=1}^\infty(a_{n}-a_{n-1})$ 
A: Yes, an easy way to see this is to look at decimal expansions, as we actually use this fact daily when we say that a number is equal to its expansion. For example,
$$\pi=3+0.1+0.04+0.001+0.0005+0.00009+0.000002+\cdots$$
$$e = 2 + 0.7 + 0.01 + 0.008 + 0.0002 +0.00008 +0.000001 +0.0000008+\cdots$$
Each partial sum is a rational number, and you can break apart any other irrational number the same way.
A: Let $\alpha$ be an irrational number. Let $n_0:=\lfloor\alpha\rfloor$ and define inductively $$n_{k+1}:=\left\lfloor10^{k+1}\left(\alpha-\sum_{i=0}^k\dfrac{n_i}{10^i}\right)\right\rfloor.$$ Then $$\sum_{k=0}^\infty\dfrac{n_k}{10^k}=\alpha.$$
Note that $\alpha$ need not be irrational.
A: Every real number can be represented as an infinite sum of rationals.
Proof: Let $a\in\mathbb{R}$ and $a_1,a_2,\dots$ be a sequence of rationals converging to $a$.
Then
$$a=a_1+\sum\limits_{n=1}^\infty(a_{n+1}-a_n)$$
A: If I understand you correctly, then the answer is no. Notice that all the sequences in question have general terms of a “regular” form. However, since the number of such “regular” expressions is countable, whereas the number of irrationals is not, the logical conclusion would be that it is simply impossible. Rob John mentioned rearranging the “regular” terms of a conditionally convergent expression to obtain every real number imaginable. True, but in this case the rearrangement itself would be “irregular”, thus disrupting the “regularity” of the expressions you gave as examples.
A: 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.
A: ====new edit====
It comes to my attention through Lucien's answer that " represented as an infinite sum of a sequence of rational numbers" can be interpreted two ways.  It can be simply $x = \sum q_n $ where each $q_n$ is a rational number.  This is the way I interpreted it and it's this interpretation that the rest of this answer is based on.
Or it could be interpreted as $x = \sum $(some nice rule that gives a rational number based on n).  The examples of the OP are of this type and have a predictive quality.  We can use them to calculate the value of the real number.  My interpretation has no predictive quality as to what the $q_n$ terms will be; just that there are a series of rational terms that will converge to the real irrational x.
By my interpretation, all irrationals can be so represented (answer below). By Lucien's interpretation, they can not.  His/Her reason is there only countably many rules.  I'm not sure of that, but I believe irrationals being uncountable make them "arbitrary" and unpredictable.  But I'd have a very difficult time formalizing that.
========== end of new edit ===========
Short answer: That is the definition of a real number.
Long answer:
The fundamental thereom of analysis is that there is an ordered field that extends the rationals such that the field has the least lower bound property.  We define the real numbers to be that field.
This means, by definition, every real number is the limit of a convergent sequence of rationals.  By definition.
Infinite sums are the limit of finite sums.  Hence every real can be written as an infinite sum of rationals. This equivalent to the definition of real number.
The proof of the fundamental thereom is kind of tedious and long. It's not hard but the point is you do the proof before the reals are defined and the definition comes out during the proof.
Longer answer:
Outline of Fund Theorem:
Step 1: Define a "cut" to be a set of rationals with properties:
i) a cut is not empty.
ii) if p is in a cut then every rational number less than p is in the cut
iii) for any p in the cut you can find a larger rational that is in the cut
So a cut could be all the rationals less than but not equal to 3.  Or all rational numbers whose squares are less than 2. (The first is going to eventually be equivalent to 3, and the latter is going to eventually be equivalent to $\sqrt 2$
Step 2: Define a < b to mean the cut a is a subset of the cut b.
Step 3: Show that the set of all cuts, let's call it R~ has the least upper bound property.
Sheesh.  This is where it gets abstract.  The least upper bound property means every bounded set in a Universal Set (such as what the Reals will be once we define them) has a distinct limit that is in the universal set.  Example: Q does not have the least upper bound property.
So we can have a set of cuts called A. It can be bounded above meaning the is a cut, b, such that all the cuts in A are subsets of b. (Remember "smaller" means "is a subset of").  The union of all the cuts in A is bigger or equal to  all cuts in A.  the union is a cut itself.  The union it the smallest cut that is bigger than all the cuts in A. So the union is a least upper bound and R~ has the least upper bound property.
Step 4: Define cut a "+" cut b to be the cut that contains the sums of elements from a plus elements from b.  Define 0~ to be the cut that contains all the negative numbers.  This satisfies addition properties.
Step 5: More about field and additive and order properties than you'd care to think about.
Step 6-8: Show that R~ a field.
Step 9: Show the Q~ = all the cuts that are defined to be all points less than a rational number is equivalent to Q. so Q has an extension that is equivalent to R~.  We call the R, the real numbers.
So.....  So each real number is simply the limit of all the rational numbers in some cut.  The cut provides sequences of rational numbers that converge to that real number. 
So every real number is the limit of a convergent sequence of rational numbers.
A: Yes, that depends on the fact that $\mathbb Q$ is dense in $\mathbb R$ and closed under subtraction. For a dense set $D$ in $X$ that is also closed under subtraction then any $x\in X$ can be expressed as a sum of elements in $D$.
Proof: Since $D$ is dense in $X$ we have for each element $x\in X$ that $x = \lim_{n\to\infty}d_n$. Since $D$ is closed under subtraction $\delta_n = d_{n}-d_{n-1} \in D$ and $d_n = d_0+\sum_1^n\delta_n$, so $x = d_0+\sum_1^\infty\delta_n$.
A: First, every rational $p/q$ can be represented as the limit of a series of rational numbers. The simplest is $a_0 = \frac{p}{q}$, $a_k = O, \forall k \neq 0$. You can do that in an infine number of ways: $a_0 = \frac{p-1}{q}$, $a_1= \frac{1}{q}$, and others $a_k$ are $0$, and you can build your own easily.
Continued fractions are often considered as  more "mathematically natural" representations of any real number than other representations such as decimal representations. You can produce a series of "best" rational approximations to any read number $\alpha$, in the shape of:
$$\alpha \sim b_0 + \frac{1}{b_1+\frac{1}{b_2+\frac{1}{b_3+\ldots}}}\,. $$
Those representations have great properties. And of course as they become rational when you stop at $b_k$, the above property for rationals holds. 
So yes, any real is the limit of an infinite quantity of sums of rational series.
A: 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 
$$0 < y < x $$
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 < 0.)
QED
