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I am trying to prove that indeed $\pi=\pi$. More precisely, that:

$$6\sum_{n=0}^\infty \frac{(2n)!}{4^n (n!)^2 (2n+1) 2^{2n+1}}=\pi$$

Where the definition of $\pi:$

$$\pi=4\sum_{n=1}^\infty\frac{(-1)^{(n+1)}}{(2n-1)}$$

Using the epsilon delta definition, we should prove:

$$\forall \epsilon_+\exists\delta\forall k(k>\delta\rightarrow|3+6\sum_{n=1}^k \frac{(2n)!}{4^n (n!)^2 (2n+1) 2^{2n+1}}-4\sum_{n=1}^k\frac{(-1)^{(n+1)}}{(2n-1)}|<\epsilon)$$

Re-arranging the sum, it can be written as:

$$\sum_{n=1}^k\frac{(-1)^{(n+1)} 4^{n+1} (n!)^2 (2n+1) 2^{(2n+1)}-6(2n)!(2n-1)}{(4n^2-1) 4^n (n!)^2 2^{2n+1} }-3$$

But from this point it's difficult to go on and find delta as a function of epsilon. What techniques should be used to prove the identity? Looking at a graph I think it's clear that $\delta=\text{ceil}(\frac{1} {\epsilon})$ would satisfy the criteria, but proving it is a other matter.

Dole
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    What you're trying to prove and your $\epsilon$-$\delta$ statement are not the same thing. The two series expressions for $\pi$ need not converge at the same speed. In fact, there are techniques known as series acceleration used to, well, accelerate speed of convergence of a given series. – Fimpellizzeri Jul 26 '18 at 19:51
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    The way you are approaching this problem is very unusual. It is customary to use the geometric/trigonometric definition (area of a unit circle, or one half the circumference) of $\pi$ as the base definition and then show that each of these series converge to that value of $\pi$. – Hamed Jul 26 '18 at 19:54
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    The sum $ \sum_{n=0}^\infty \binom{2n}{n}8^{-n}/(2n+1)=\pi/(2\sqrt{2}) $ according to Mathematica and not the value $\pi /3$ as implied in the statement of the problem. – user321120 Jul 26 '18 at 20:15
  • You could try $2 \sum _{n=0}^{\infty } \frac{1 }{4^n (2 n+1)}\binom{2 n}{n}=\pi$ instead perhaps. – James Arathoon Jul 26 '18 at 20:22
  • @skbmoore More specifically, $$\sum_{n=0}^\infty \binom{2n}{n} \frac{x^n}{2n+1} = \frac{\arcsin(2\sqrt{x})}{2\sqrt{x}}$$ for $\lvert x\rvert < 1/4$. – Clement C. Jul 26 '18 at 20:23
  • @skbmoore Oops, typo!! Corrected. – Dole Jul 26 '18 at 20:24
  • Some amateur many years ago was claiming that whenever two series of rationals converge to the same irrational sum, that it should be provable by some series of manipulations like grouping, rearranging, and so on. So, a more difficult version of this problem would be to exhibit such a simple proof for this. – GEdgar Jul 26 '18 at 20:30
  • @GEdgar Well, a posteriori this seems rather obvious (regardless of whether the series are of rationals) using a 'sum until this point' kind of argument, like one uses when showing that conditionally convergent series can be rearranged to result in whatever you want. – Fimpellizzeri Jul 26 '18 at 21:53
  • @GEdgar What if it is true but unprovable? Hehe... Anyway, I managed to prove by induction that the sum is smaller than $1/n$. If the sum could be put into a difference equation form and then solved, it would be easy to solve of course... but I could not come up with any equations or smart re-groupings. Maybe some other amateur could ;). – Dole Jul 27 '18 at 01:01

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Consider the expression \begin{eqnarray*} \sum_{n=0}^{\infty} \binom{2n}{n} \frac{x^{2n+1}}{2n+1}. \end{eqnarray*} Now use \begin{eqnarray*} \frac{x^{2n+1}}{2n+1}= \int_0^x x^{2n} dx \end{eqnarray*} and \begin{eqnarray*} \sum_{n=0}^{\infty} \binom{2n}{n} x^{2n} = \frac{1}{\sqrt{1-4x^2}}. \end{eqnarray*} Invert the order of the plum & integral and the expression becomes \begin{eqnarray*} \int_0^x \frac{1}{\sqrt{1-4x^2}} dx. \end{eqnarray*} This integral can be done by the substition $2x=\sin( \theta)$ to give \begin{eqnarray*} \frac{1}{2}\sin^{-1} (2x). \end{eqnarray*} Now substitute $x^2=1/8$ and we have \begin{eqnarray*} \sum_{n=0}^{\infty} \frac{1}{2n+1}\binom{2n}{n} \frac{1}{8^{n}}= \frac{\pi}{2 \sqrt{2}}. \end{eqnarray*}

Donald Splutterwit
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