A couple of formulas for $\pi$

Question

While I was studying sums of polygonal numbers I discovered a couple of formulas for $\pi$. Most of the formulas I found were already known, but I can't seem to find any references to the following four: $$\pi=\sum_{n=1}^\infty \frac{3}{n(2n-1)(4n-3)},$$ $$\pi=\sum_{n=1}^\infty \frac{3\sqrt{3}}{(3n-1)(3n-2)},$$ $$\pi=4\sqrt{3}\sum_{n=1}^\infty \frac{12n-5}{8n(2n-1)(3n-1)(6n-5)},$$ $$\pi=16\sum_{n=1}^\infty \frac{864n(n-1)+226}{(12n-1)(12n-5)(12n-7)(12n-11)(4n-1)(4n-3)}.$$ Are the above formulas known? I have looked here and here. Because of their simplicity, I find it hard to believe that the first two formulas are unknown.

Answer 1

The second formula is known from Fourier analysis, and perhaps that is a place to look for the other formulas as well. The Fourier series of $f(x)={\pi-x \over 2}$ is given by $${\pi-x \over 2}=\sum_{n=1}^\infty {\sin(nx)\over n}, \qquad 0<x<2\pi$$ Let $x={2\pi\over 3}$, and you get the second formula above.

Addendum: The third formula can be shown in the same way as the second. By the method of partial fractions, the sum can be rewritten as $$4\sqrt3 \sum_{n=1}^\infty {12n-5 \over 8n(2n-1)(3n-1)(6n-5)}=3\sqrt3 \sum_{n=1}^\infty \Bigl({1\over 6n}-{1\over 6n-2}-{1\over 6n-3}+{1\over 6n-5}\Bigr)$$ The last expression is equal to $$6\sum_{n=1}^\infty {\sin\bigl((n+1){\pi\over 3}\bigr)\over n}$$ which can be summed to yield the result $\pi$.

One way is to do this is to expand the nominator $\sin((n+1){\pi\over 3})={1\over 2}\sin({n\pi\over 3})+{\sqrt3\over 3}\cos({n\pi\over 3})$ and to use the formulas $\sum_{n=1}^\infty{\sin(nx)\over n}={\pi-x\over 2}$ and $\sum_{n=1}^\infty{\cos(nx)\over n}=\log(2-2\cos(x))$, which are valid for $0<x<2\pi$, and to evaluate these expressions at $x={\pi\over 3}$.

Addendum We now derive the two remaining formulas. Partial fractions on the first formula yields $$\sum_{n=1}^\infty{3\over n(2n-1)(4n-3)}=4\sum_{n=1}^\infty\Bigl({1\over 4n}-{3\over 4n-2}+{2\over 4n-3}\Bigr)$$ Here we need more than one trigonometric function, but it is possible to rewrite this as $$4\sum_{n=1}^\infty {\sin(n{\pi\over 2})+2\cos(n{\pi\over 2})-\cos(n\pi) \over n}=4\Bigl(f\Bigl({\pi\over 2}\Bigr)+2g\Bigl({\pi\over 2}\Bigr)-g(\pi)\Bigr) = 4\Bigl({\pi\over 4}+8\log(2)-4\log(4)\Bigr)=\pi$$ where $f(x)={\pi-x\over 2}$ and $g(x)=-{1\over 2}\log(2-2\cos(x))$, as above.

Finally, partial fractions on the fourth formula yields $$16\sum_{n=1}^\infty {864n(n-1)+226\over (12n-1)(12n-5)(12n-7)(12n-11)(4n-1)(4n-3)}$$ $$=6\sum_{n=1}^\infty\Bigl(-{1\over 12n-1}+{2\over 12n-3}-{1\over 12n-5}+{1\over 12n-7}-{2\over 12n-9}+{1\over 12n-11}\Bigr) $$ $$= 6\sum_{n=1}^\infty{-{1\over 3}\sin{n\pi\over 6}+{4\over 3}\sin{3n\pi\over 6}-{1\over 3}\sin{5n\pi\over 6} \over n}$$ $$=6\Bigl(-{1\over 3}f\Bigl({\pi\over 6}\Bigr)+{4\over 3}f\Bigl({3\pi\over 6}\Bigr)-{1\over 3}f\Bigl({5\pi\over 6}\Bigr)\Bigr)=6\Bigl(-{1\over 3}{5\pi\over 12}+{4\over 3}{\pi\over 4}-{1\over 3}{\pi\over 12}\Bigr)=\pi$$ where we still have $f(x)={\pi-x\over 2}$.

Addendum: The Fourier series of a function $f(x)$ defined on the interval $[0,2\pi]$ is given by $$a_0+\sum_{n=1}^\infty\bigl(a_n\cos(nx)+b_n\sin(nx)\bigr)$$ where $$a_0={1\over 2\pi}\int_0^{2\pi}f(x){\rm d}x,\qquad a_n={1\over \pi}\int_0^{2\pi}f(x)\cos(nx){\rm d}x,\qquad b_n={1\over \pi}\int_0^{2\pi}f(x)\sin(nx){\rm d}x$$ In our case, we get $a_n=0$ for all $n\geq 0$ from the symmetry of $f(x)$ about the point $x=\pi$ (i.e., $f(\pi-x)=-f(\pi+x)$), and we get $b_n={1\over n}$ by integration by parts. The Fourier series will converge to $f(x)$ at all points where (the periodic extension of) $f$ is differentiable.