Trigonometric way to find the solution of Basel problem

Let's generalise the whole thing first

Generally \[\sum_{n=1}^∞ \dfrac{\sin(n\theta)}{n} = \dfrac{π-\theta}{2} \] Where \(0<\theta≤π\)

And consequently ∑n=1∞cos⁡(nθ)n2=∑n=1∞1n2−πθ2+θ24=ζ(2)−πθ2+θ24\sum_{n=1}^∞ \dfrac{\cos(n\theta)}{n^2} = \sum_{n=1}^∞ \dfrac{1}{n^2} -\dfrac{π\theta}{2} +\dfrac{\theta^2}{4} =\zeta(2)-\dfrac{π\theta}{2} +\dfrac{\theta^2}{4} refer this solution

let's take this sum ∑n=1∞sin⁡2(nθ)n2\displaystyle\sum_{n=1}^∞ \dfrac{\sin^2(n\theta)}{n^2} It is =∑n=1∞(12n2−cos⁡(2nθ)2n2)=\sum_{n=1}^∞\left( \dfrac{1}{2n^2} - \dfrac{\cos(2n\theta)}{2n^2} \right) =ζ(2)2−(ζ(2)2+(2θ)28−πθ2)= \dfrac{\zeta(2)}{2} -\left (\dfrac{\zeta(2)}{2} +\dfrac{(2\theta)^2}{8} -\dfrac{π\theta}{2}\right ) =θ(π−θ)2= \dfrac{\theta(π-\theta)} {2} Now ∑n=1∞sin⁡2(nπ2)n2=π2(π−π/2)2=π28\sum_{n=1}^∞ \dfrac{\sin^2(n\frac{π}{2})}{n^2} = \dfrac{ \frac{π}{2} (π-π/2)}{2} = \dfrac{π^2}{8}   ⟹  112+132+152+⋯=π28\implies \dfrac{1}{1^2} +\dfrac{1}{3^2} +\dfrac{1}{5^2} +\cdots= \dfrac{π^2}{8} But this is all odd terms . ζ(2)4+∑n=0∞1(2n+1)2=ζ(2)\dfrac{\zeta(2)}{4} +\sum_{n=0}^∞\dfrac{1}{(2n+1)^2} = \zeta(2) ζ(2)=43×π28=π26\zeta(2)= \dfrac{4}{3} × \dfrac{π^2}{8} = \dfrac{π^2}{6}

#Calculus

Note by Dwaipayan Shikari
2 months, 1 week ago

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