Since $$\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}$$

can we now express $\pi$ in terms of this series by multiplying by $6$ and taking the square root? If not why is this not true?

I was wondering since I had an exam question that required to write $\pi$ in terms of some infinite sum. I did it exactly like this and got 0 points. So I thought maybe I'm doing something wrong by manipulating it this way

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    $\begingroup$ Sure, $\pi=\sqrt{6\sum_{n=1}^\infty \frac 1{n^2}}$... Or are you asking a different question? Note that this is a very slow way to calculate the digits of $\pi$... $\endgroup$
    – abiessu
    May 29, 2018 at 13:00
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    $\begingroup$ Out of context, it's difficult to guess what the grader was thinking. Perhaps there was some detail of the question that would force you to use a different series, or perhaps there was no such detail and the grader just wasn't aware of this series. It might be worth posting the question exactly as it was asked, or maybe just ask the instructor or TA. $\endgroup$
    – David K
    May 29, 2018 at 13:15
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    $\begingroup$ As written, you have not written it as an infinite sum but as a square root of an infinite sum. A bit picky mind! $\endgroup$
    – Paul
    May 29, 2018 at 13:44
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    $\begingroup$ @Paul OP said in terms of an infinite sum. Wouldn't you regard expressing $\sqrt x$ as a function of $x$? $\endgroup$
    – TheSimpliFire
    May 29, 2018 at 13:59
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    $\begingroup$ Illegal seems to be a very wrong word. $\endgroup$ May 29, 2018 at 14:18

3 Answers 3


The resulting formula for $\pi$ is certainly correct; but depending on the question's original wording, "in terms of" a sum might have meant as a sum, not a function thereof (such as its square root). Another option is the Gregory series $\pi=4\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}$, which follows from the Taylor series of $\arctan x$ evaluated at $x=1$.

  • $\begingroup$ I was really just more curious whether this was correct. Don't care much about how many points I got. Although I do admit I was most likely wrong using this specific series cause after all it's not very easy to prove that its sum indeed is equal to $\frac{\pi^2}{6}$. $\endgroup$
    – DreaDk
    May 29, 2018 at 14:07
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    $\begingroup$ It's worth mentioning that the Gregory series is in practice a much worse method of calculating $\pi$, since it converges pathetically slow. $\endgroup$ May 29, 2018 at 15:32
  • $\begingroup$ ...actually, $\sum_{n=0}^\infty \frac1{n^2}$ doesn't converge faster either though, so... $\endgroup$ May 29, 2018 at 15:39
  • $\begingroup$ See math.stackexchange.com/questions/14113/… $\endgroup$
    – J.G.
    May 29, 2018 at 15:43

Yes, of course. $$\pi=\sqrt{6\left(1+\frac14+\frac19+\cdots\right)}$$

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    $\begingroup$ Yes, yes, I'm going to take the square root. But first I have to multiply by 6. And before I can multiply, I have to finish adding up all these reciprocal squares. I've been at it for a while, I'm sure I'm almost done. $\endgroup$
    – corsiKa
    May 29, 2018 at 17:30
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    $\begingroup$ @corsiKa: The trick is to do each step in half the time of the previous one. $\endgroup$
    – celtschk
    May 29, 2018 at 19:17

You can use the unique factorization properties of Gaussian integers to prove the familiar sum

$\frac{\pi}{4}= 1 - (1/3) + (1/5) - (1/7) + ...$

See here for the connection between this and UF of Gaussian integers.

Less well known is the fact that we can use any other UF domain of imaginary quadratic surds, and with a proof analogous to the one in the referenced video we get things like:

$m+n\sqrt{-2}$ domain:

$\frac{\pi}{2\sqrt{2}}= 1 + (1/3) - (1/5) - (1/7) + (1/9) + (1/11) - (1/13) - (1/15) + ...$

(Same terms as the $\pi/4$ series but a $++--$ sign pattern.)

$m+n\omega$ domain, $\omega$ is a primitive cube root of unity:

$\frac{\pi}{3\sqrt{3}}= 1 - (1/2) + (1/4) - (1/5) + (1/7) - (1/8) + (1/10) - (1/11) + ... $

(Skip multiples of 3 and then alternate signs.)

$m+n(1+\sqrt{-7})/2$ domain:

$\frac{\pi}{\sqrt{7}}= 1 + (1/2) - (1/3) + (1/4) - (1/5) - (1/6) + (1/8) + (1/9) - (1/10) + (1/11) - (1/12) - (1/13) + ...$

(Skip multiples of 7 and apply the sign pattern $++-+--$ to the rest.)

We can keep on going all the way to the $\sqrt{-163}$ domain, but the sign patterns (determined by Legendre symbols) get more and more complicated.

  • 1
    $\begingroup$ Appreciate the connections made but it's a bit out of my level of knowledge right now. Will keep an eye on it. $\endgroup$
    – DreaDk
    May 29, 2018 at 14:43

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