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