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.

  • $\begingroup$ How did you discover these, may I ask? $\endgroup$ – ShreevatsaR Nov 15 '12 at 6:37
  • $\begingroup$ @ShreevatsaR: By experimenting with series of polygonal numbers, for example, the third one was originally obtained by: $$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{p_3(n)+p_9(n)}=\frac{\pi}{4\sqrt{3}},$$ where $p_m(k)$ is the $k$th $(m+2)$-gonal number. $\endgroup$ – Carolus Nov 15 '12 at 7:03
  • $\begingroup$ There seem to be an abundance of these formulas, I found two more yesterday, but I bet you could find ten or twenty in a day if you put your mind to it ;) $\endgroup$ – Carolus Nov 15 '12 at 7:06
  • $\begingroup$ I'm afraid I still don't understand: how did you get that equality in the first place? (Did you write down the sum of $\frac{(-1)^{n-1}}{p_3(n)+p_9(n)}$, and happen to just guess that it is $\frac{\pi}{4\sqrt{3}}$?) $\endgroup$ – ShreevatsaR Nov 16 '12 at 9:23
  • $\begingroup$ @ShreevatsaR: Well, yes. I was looking for something else, a simple closed form for some sum involving polygonal numbers (using Mathematica). Then $\pi$ started popping up again and again in what I thought were strange circumstances. Then it was just a matter of modifying the formulas to find nicer expressions. Sorry if I am explaining poorly. $\endgroup$ – Carolus Nov 16 '12 at 9:54

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.

  • $\begingroup$ This is very nice. might you add how to turn the sin into polynomials in that example for my benefit? $\endgroup$ – sperners lemma Nov 12 '12 at 16:11
  • 1
    $\begingroup$ @spernerslemma I've added the main steps. $\endgroup$ – Per Manne Nov 12 '12 at 16:46
  • $\begingroup$ thanks that's really cool. $\endgroup$ – sperners lemma Nov 12 '12 at 16:47
  • $\begingroup$ Wow, this is great! I should definitely do some reading on Fourier analysis. Do you think the fourth formula can be derived similarly? $\endgroup$ – Carolus Nov 13 '12 at 12:57
  • $\begingroup$ @Carolus I've added the main steps of the last two formulas. There were some work in deriving the right coefficients for the fourth formula, but the method is straight forward, and you only need persistance to get the final result. $\endgroup$ – Per Manne Nov 14 '12 at 0:02

The first formula may be rewritten by changing the lower limit of the summation:

$$\begin{align} \frac{\pi}{3} &=\sum_{n=1}^\infty \frac{1}{n(2n-1)(4n-3)} \\ &=\sum_{n=0}^\infty \frac{1}{(n+1)(2(n+1)-1)(4(n+1)-3)} \\ &=\sum_{n=0}^\infty \frac{1}{(n+1)(2n+1)(4n+1)} \\ \end{align}$$

This last equation is given by Lehmer in page 139 here.

A similar one that uses four factors is $$\pi-3=\sum_{k=0}^\infty \frac{24}{(4k+2)(4k+3)(4k+5)(4k+6)}$$

(see Series and integrals for inequalities and approximations to $\pi$)

A formula which is similar to the third one you give is: $$\pi=72 \sqrt{3} \sum_{k=0}^\infty \frac{1}{(6 k+1) (6 k+2) (6 k+4) (6 k+5)}$$ which has all terms positive and constant numerator. Taking the first term of the series out of the summation, this proves $\pi>\frac{9}{5}\sqrt{3}\approx 3.12$.

There is also $$\pi=\frac{23040(17+9 \sqrt{3})}{23} \sum_{k=0}^\infty \frac{1}{(12 k+1) (12 k+3) (12 k+5) (12 k+7) (12 k+9) (12 k+11)}$$ with the denominators of your fourth formula. The first term in this series leads to the approximation

$$\pi \approx \frac{2^9(17+9\sqrt{3})}{5313}=\frac{2^{10}((2^4+1)+(2^3+1)\sqrt{3})}{21·22·23} = \frac{2^{10}}{231(17-9\sqrt{3})}\approx 3.140$$

Using $163$ instead of $\frac{5313}{17+9\sqrt{3}} \approx 163.033$ gets one more correct digit. The approximation $$\pi \approx \frac{2^9}{163} \approx 3.1411$$ is due to Stoschek (see http://mathworld.wolfram.com/PiApproximations.html)

Also compare to $\pi \approx \frac{2^{13}}{3(-383+560\sqrt{5})}$ given by T.Piezas here.


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