Is it possible to represent every irrational number as a (limit of) an infinite sum of rational numbers? [duplicate]

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.

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.

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

$$a=a_1+\sum\limits_{n=1}^\infty(a_{n+1}-a_n)$$

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.