5
$\begingroup$

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.

$\endgroup$
9
  • 6
    $\begingroup$ 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. $\endgroup$ Jul 26, 2018 at 19:51
  • 1
    $\begingroup$ 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 $\pi$. $\endgroup$
    – Hamed
    Jul 26, 2018 at 19:54
  • 2
    $\begingroup$ 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. $\endgroup$
    – user321120
    Jul 26, 2018 at 20:15
  • $\begingroup$ You could try $2 \sum _{n=0}^{\infty } \frac{1 }{4^n (2 n+1)}\binom{2 n}{n}=\pi$ instead perhaps. $\endgroup$ Jul 26, 2018 at 20:22
  • $\begingroup$ @skbmoore More specifically, $$\sum_{n=0}^\infty \binom{2n}{n} \frac{x^n}{2n+1} = \frac{\arcsin(2\sqrt{x})}{2\sqrt{x}}$$ for $\lvert x\rvert < 1/4$. $\endgroup$
    – Clement C.
    Jul 26, 2018 at 20:23

1 Answer 1

2
$\begingroup$

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*}

$\endgroup$
1
  • 6
    $\begingroup$ Do you have a reference for the theorem on plum and integral swapping? :) $\endgroup$
    – Clement C.
    Jul 26, 2018 at 20:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .