Is this formula for $\pi$ already known?

I found that

$$\pi = \frac{10}{3}-2\sum_{n=1}^{\infty}\frac{(2n)!}{(2n^2+5n+3)\cdot (2^n\cdot n!)^2}$$

Is this already known? If so, where might I find it?

EDIT: Here's how I found the formula, since someone asked:

We start with an equation describing the unit circle, $y^2+x^2=1$. We aim to find the area one quadrant to give us the value of $\frac{\pi}{4}$.

Note that this is equivalent to evaluating: $$\int_0^1\sqrt{1-x^2}dx$$

Expanding $(1-x^2)^{\frac{1}{2}}$ by the binomial theorem yields:

$$1+\frac{1}{2}\cdot(-x^2)+\frac{\frac{1}{2}\cdot\frac{-1}{2}}{2!}\cdot(-x^2)^2+\frac{\frac{1}{2}\cdot\frac{-1}{2}\cdot\frac{-3}{2}}{3!}\cdot(-x^2)^3+\frac{\frac{1}{2}\cdot\frac{-1}{2}\cdot\frac{-3}{2}\cdot\frac{-5}{2}}{4!}\cdot(-x^2)^4\space...$$

$$=1-\frac{x^2}{2}+\frac{1}{2}\left(\frac{\frac{-1}{2}}{2!}\cdot(-x^2)^2+\frac{\frac{-1}{2}\cdot\frac{-3}{2}}{3!}\cdot(-x^2)^3+\frac{\frac{-1}{2}\cdot\frac{-3}{2}\cdot\frac{-5}{2}}{4!}\cdot(-x^2)^4\space...\right)$$

$$=1-\frac{x^2}{2}+\frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^n\cdot\prod_{k=1}^n(2k-1)}{2^n\cdot(n+1)!}\cdot (-1)^{n+1} x^{2(n+1)}$$

First, note that $(-1)^n\cdot(-1)^{n+1}=(-1)^{2n+1}=-1$, hence: $$=1-\frac{x^2}{2}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{\prod_{k=1}^n(2k-1)}{2^n\cdot(n+1)!}\cdot x^{2(n+1)}$$

Now we will attempt to reduce $\prod_{k=1}^n(2k-1)$, the product of the first $n$ odd integers:

$$\prod_{k=1}^n(2k-1)=1\cdot 3\cdot 5\cdot 7\cdot\space ...\space\cdot (2n-1)$$

$$=\frac{1\cdot 2\cdot 3\cdot 4\cdot\space ...\space\cdot(2n)}{2\cdot 4\cdot 6\cdot 8\cdot\space ...\space\cdot(2n)}$$

$$=\frac{(2n)!}{2(1)\cdot 2(2)\cdot 2(3)\cdot 2(4)\cdot\space ...\space\cdot2(n)}$$

$$=\frac{(2n)!}{2^n\cdot n!}$$

This gives us: $$\sqrt{1-x^2}=1-\frac{x^2}{2}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{(2n)!}{2^n\cdot n!\cdot2^n\cdot(n+1)!}\cdot x^{2(n+1)}$$

$$=1-\frac{x^2}{2}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{(2n)!\cdot x^{2(n+1)}}{(n+1)\cdot\left(2^n\cdot n!\right)^2}$$

Now, we look to find:

$$\int_0^1\left(1-\frac{x^2}{2}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{(2n)!\cdot x^{2(n+1)}}{(n+1)\cdot\left(2^n\cdot n!\right)^2}\right)dx$$

$$=\left[x-\frac{x^3}{6}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{(2n)!\cdot x^{2n+3}}{(2n+3)\cdot(n+1)\cdot\left(2^n\cdot n!\right)^2}\right]_0^1$$

$$=\left[\frac{5}{6}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{(2n)!}{(2n^2+5n+3)\cdot\left(2^n\cdot n!\right)^2}\right]-[0]$$

Since we know that: $$\frac{\pi}{4}=\frac{5}{6}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{(2n)!}{(2n^2+5n+3)\cdot\left(2^n\cdot n!\right)^2}$$

We can find an expression for $\pi$:

$$\pi=\frac{10}{3}-2\left(\sum_{n=1}^{\infty}\frac{(2n)!}{(2n^2+5n+3)\cdot\left(2^n\cdot n!\right)^2}\right)$$

Computation of the first few terms gives an approximate value of $\pi$:

$$\pi\approx\frac{10}{3}-2\left(\frac{2}{10\cdot 4}+\frac{24}{21\cdot 64}+\frac{720}{36\cdot 2304}+\frac{40320}{55\cdot 147456}+\frac{3628800}{78\cdot 14745600}\right)$$

$$\approx 3.164004659$$

Solution 1:

I don't know if it is known, but it it is not hard to derive. First note that $2n^2+5n+3=(n+1)(2n+3).$ Then let:

$$g(z)=\sum_{n=1}^{\infty} \frac{1}{(n+1)(2n+3)}\binom{2n}{n}z^{2n+3}.$$ Your expression is then: $$\frac{10}3-16g\left(\frac12\right).$$

Then we see:$$g''(z)=2\sum_{n=1}^{\infty}\binom{2n}{n}z^{2n+1}=\frac{2z}{\sqrt{1-4z^2}} -2z$$

Calculus techniques let us integrate the expression $\frac{z}{\sqrt{1-4z^2}}$ twice and get the correct constant terms to get a value for $g\left(\frac12\right).$