The following series converge to a value relating to $\pi$: \begin{align} \frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots&=\frac{\pi}{4},\\ \frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots&=\frac{\pi^2}{8},\\ \frac{1}{1^3}-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+\cdots&=\frac{\pi^3}{32},\\ \frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+\frac{1}{7^4}+\cdots&=\frac{\pi^4}{96},\\ \frac{1}{1^5}-\frac{1}{3^5}+\frac{1}{5^5}-\frac{1}{7^5}+\cdots&=\frac{5\pi^5}{1536}. \end{align}

It seems that if we define $$f(n):=\sum_{i=0}^{\infty}\Big(\frac{(-1)^i}{2i+1}\Big)^n,\quad n\in\mathbb{N}_+,$$ then the values of $f$ are related to $\pi$, and in fact I guess we have $$f(n)=A\pi^n,\quad A\in\mathbb{Q}.$$

This is strongly reminiscent of Basel problem, where we have a famous solution based on the Weierstrass factorization theorem. Trying to imitate that proof, I want to find a real function $g$ with $$Z:=g^{-1}(0)=\Big\{\frac{(-1)^i}{2i+1}:i\in\mathbb{N}\Big\},$$ and $g$ can be factorized as something like $$g(x)=\prod_{a\in Z}\Big(1-\frac{x}{a}\Big),$$ and by comparing the Taylor series of $g$ and applying Vieta's formulas and Newton's identities, we might find the value of $f(1)$ or even more. But these are just wild guesses. I haven't even studied complex analysis, and I'm only imitating the proof for Basel problem. I wonder if this leads to any reasonable solution.

My question is: How do we obtain the value of $f(n)$ and how do we prove these? Don't hesitate to post solutions based on complex analysis or more advanced analysis! Thanks in advance.


This method is overkill, but here goes.

For even $n$, $$\sum_{k=0}^\infty\frac1{(2k+1)^n}=\left(1-\frac1{2^n}\right)\zeta(n).$$ The value of $\zeta(n)$ for even $n$ is well-known. It can be obtained from the functional equation connecting $\zeta(s)$ and $\zeta(1-s)$.

For odd $n$, $$\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^n}=L(n,\chi)$$ where $\chi$ is a Dirichlet character of conductor $4$, and $L$ is a Dirichlet L-function There is a functional equation connecting $L(s,\chi)$ and $L(1-s,\chi)$. One can compute $L(n,\chi)$ for odd positive $n$ by using this.

Of course there are more elementary ways of obtaining both these results.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.