I came across the set of problems
$$ \begin{split} \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} &= \frac11 &- \frac13 &+ \frac15 &\pm \ldots &= \frac{\pi}{4}\\ \sum_{k=0}^\infty \frac{1}{(2k+1)^2} &= \frac1{1^2} &+ \frac1{3^2} &+ \frac1{5^2} &+ \ldots &= \frac{\pi^2}{8}\\ \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^3} &= \frac1{1^3} &- \frac1{3^3} &+ \frac1{5^2} &\pm \ldots &= \frac{\pi^3}{32}\\ \sum_{k=0}^\infty \frac{1}{(2k+1)^4} &= \frac1{1^4} &+ \frac1{3^4} &+ \frac1{5^4} &+ \ldots &= \frac{\pi^4}{96}\\ \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^5} &= \frac1{1^5} &- \frac1{3^5} &+ \frac1{5^5} &\pm \ldots &= \frac{5\pi^5}{1536}\\ \sum_{k=0}^\infty \frac{1}{(2k+1)^6} &= \frac1{1^6} &+ \frac1{3^6} &+ \frac1{5^6} &+ \ldots &= \frac{\pi^6}{960} \end{split} $$
I was able to do the first one using Maclaurin series of $\tan x$ , and also I was able to compute the ones with even powers by comparing coefficient of various powers of x in two different expressions for $\sin x$ , one from Maclaurin expansion and the other by forming a polynomial with roots {${...,-2\pi,-\pi,0,\pi,2\pi,...}$} (i.e. Euler's Method)
But I have no idea how to compute the ones with odd powers , I searched a lot too , but nowhere could I find any way to compute the ones with odd powers.
Could someone please guide me on how to compute the ones with odd powers ?
And also shorter ways of computing the ones with even powers are highly welcomed.
Thanks
