Integer based series evaluating to $\pi$

Question

Evaluate $$\sum_{n\geqslant0}\frac{\binom{2n}{n}}{2^{4n}(2n+1)}-\frac{1}{\pi}\sum_{n\geqslant1}\frac{1}{n^2\binom{2n}{n}}$$

At the first sight, I thought that this expression cannot have an elementary closed form. But then I used a calculator and something strange or rather miraculous happened. Both the terms above have very neat and clean closed form or rather both evaluate to a real number. The first term equals $\frac{\pi}{3}$ and the second term equals $\frac{\pi^2}{18}$

How can a series, that is integer based, evaluate to an irrational number and that too $\pi$. I'm having a hard time digesting this fact let alone proving the result. Any help is greatly appreciated.

Edit: I understood that integer based series can take irrational values in many cases. But still I'm not able to prove the results that the calculator gave me. Please guide me.

Answer 1

What definition do you know for real numbers?

Whatever the definition, it boils down to: $\mathbb{R}$ (set of real numbers) is defined as the completion of $\mathbb{Q}$ (set of rational numbers).

Usually this is defined as: every real number $r$ is an equivalence class of Cauchy sequences of rational numbers (there is an alternate definition with Dedekind cuts, but it is less used). Intuitively, taking equivalence classes is needed because there is an infinity of sequences that converge to any real number, and we do not want to distinguish between them.

You may ignore what a Cauchy sequence is, and/or what an equivalence class is. So let's be simpler. Any real number can be defined (not uniquely) by a sequence of rational numbers $(a_n)_{n=1 \dots \infty}$ that verifies the following property: for any $\varepsilon > 0$ we can find an $N$ for which all $a_n$ with $n \ge N$ are contained in an interval of length $\le \varepsilon$.

Now let's define sequence $(b_n)_{n=1 \dots \infty}$ by:
$b_0=a_0$, and $b_n = a_n-a_{n-1}$ for $n \ge 1$.
Then $a_n = \sum_{k=0}^n b_k$
so if $(a_n)$ is a sequence of rational numbers that converges to a real number $r$, $(b_n)$ is a series of rational numbers that sums to the same real number $r$.

So given a real number $r$, it is always possible to find a series of rational numbers whose sum is $r$.

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Note: for specific sequence (and hence, series) of rational numbers that converges to a specific real number, you may want to read about continued fractions. The sequence elements are the approximations for a given real number with the smallest possible denominators.

Answer 2

From the series expansion for $\arcsin$, we have

$$\frac{\arcsin(x)}x = \sum_{n=0}^\infty \frac{\binom{2n}n}{2n+1} \left(\frac x2\right)^{2n}$$

and from its square,

$$2\arcsin^2(x) = \sum_{n=0}^\infty \frac{1}{n^2\binom{2n}n} \left(2x\right)^{2n}$$

Both series converge for $x\in[-1,1]$; pick the right value(s) to recover the sums in question.