Solve Convergent Series with Mathematica: Pi Squared/8

[Hertz]

Hi, Mathematica is telling me the value of this series, but I can't figure out how to do it on paper. Can someone please explain? $$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$

 

[mathman]

Hi, Mathematica is telling me the value of this series, but I can't figure out how to do it on paper. Can someone please explain? $$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$

http://ptrow.com/articles/Infinite_Series_Sept_07.htm
Above describes the derivation of
$$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$

Your equation is readily derived from this. Add all the $$\frac{1}{(2n)^2}$$ terms to your series and then subtract that series as $(\sum_{n=1}^{\infty}\frac{1}{n^2})/4$. You end up with $$\frac{\pi^2}{6}-\frac{1}{4}\frac{\pi^2}{6}=\frac{\pi^2}{8}$$

 

[Hertz]

Thanks!

 

[WWGD]

I know the proof is written in detail in Dunham's "Journey Through Genius".

 

[CellCoree]

Hi there,

Thank you for reaching out. Mathematica is a powerful tool for solving mathematical problems, including series. In this case, the series you have provided is known as the Basel problem, which was famously solved by Leonhard Euler in the 18th century. Mathematica is able to solve this series using various methods, including the Euler-Maclaurin formula and the Riemann zeta function.

To solve this series on paper, you can use the Euler-Maclaurin formula, which states that for a function f(x) and a positive integer n, the following equation holds:

$$\sum_{k=0}^{n}f(k)=\int_{0}^{n}f(x)dx+\frac{1}{2}f(n)+\sum_{j=1}^{\infty}\frac{B_{2j}}{(2j)!}\left(f^{(2j-1)}(n)-f^{(2j-1)}(0)\right)$$

where B2j are the Bernoulli numbers. In this case, we can let f(x) = 1/(2x+1)^2 and n = ∞. Using this formula, we get:

$$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\int_{0}^{\infty}\frac{1}{(2x+1)^2}dx+\frac{1}{2}\left(\frac{1}{(2\infty+1)^2}\right)-\frac{1}{2}\left(\frac{1}{(2\cdot0+1)^2}\right)+\sum_{j=1}^{\infty}\frac{B_{2j}}{(2j)!}\left(\frac{d^{2j-1}}{dx^{2j-1}}\left(\frac{1}{(2x+1)^2}\right)\bigg|_{x=\infty}-\frac{d^{2j-1}}{dx^{2j-1}}\left(\frac{1}{(2x+1)^2}\right)\bigg|_{x=0}\right)$$

The first term on the right-hand side is equal to π^2/8, and the second term is equal to 0. The remaining terms can be evaluated