Calculating the infinite sum of one over the odd numbers squared: $\sum_{i \ge 0} \frac{1}{(2i+1)^2}$

Question

Can someone tell me how to calculate the following infinite sum? $$ (1/1^2)+(1/3^2)+(1/5^2)+(1/7^2)+(1/9^2)+(1/11^2)+ \cdots $$

Don't give me the answer. Can you tell me if this is a geometric series?

Answer 1

By absolute convergence, you may write $$ \sum_{n=1}^{\infty}\frac{1}{{n}^2}=\sum_{p=1}^{\infty}\frac{1}{{(2p)}^2}+\sum_{p=1}^{\infty}\frac{1}{{(2p-1)}^2}=\frac14\sum_{p=1}^{\infty}\frac{1}{p^2}+\sum_{p=1}^{\infty}\frac{1}{{(2p-1)}^2}. $$ Can you take it from here?

Answer 2

To test if it's geometric, compare the ratio of adjacent terms. For example, $\frac{1/3^2}{1/1^2}$ and $\frac{1/5^2}{1/3^2}$. Are these two quantities equal?

Hint: $$ \left( \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \cdots\right) \left( \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{8^2} + \frac{1}{16^2} + \cdots \right) \\ = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \cdots $$

Answer 3

METHODOLGY $1$

Here, we present a way forward that does not require prior knowledge of the value of the series $\sum_{n=1}\frac{1}{n^2}=\frac{\pi^2}{6}$, the Riemann-Zeta Function, or dilogarithm function. Rather, we apply straightforward analysis that includes application of the residue theorem.

To that end, note that we can write the series of interest as

$$\begin{align} \sum_{n=0}^\infty \frac{1}{(2n+1)^2}&=\sum_{n=1}^N \int_0^1 x^{2n}\,dx\int_0^1y^{2n}\,dy\\\\ &=\int_0^1\int_0^1 \sum_{n=0}^\infty(x^2y^2)^n\,dx\,dy\\\\ &=\int_0^1\int_0^1 \frac{1}{1-x^2y^2}\,dx\,dy\\\\ &=\frac12\int_0^1\frac{\log(1+x)-\log(1-x)}{x}\,dx\tag 1 \end{align}$$

Then, we enforce the substitution $x\to \frac{x-1}{x+1}$ in $(1)$ to obtain

$$\frac12\int_0^1\frac{\log(1+x)-\log(1-x)}{x}\,dx=\int_1^\infty \frac{\log(x)}{x^2-1}\,dx\tag 2$$

Next, enforcing the substitution $x\to 1/x$ in $(2)$ reveals

$$\int_1^\infty \frac{\log(x)}{x^2-1}\,dx=\int_0^1 \frac{\log(x)}{x^2-1}\,dx \tag 3$$

Adding $(2)$ and $(3)$ and dividing by $(2)$ yields

$$\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\frac12\int_0^\infty \frac{\log(x)}{x^2-1}\,dx$$

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Moving to the complex plane, we evaluate the integral $J$ defined by

$$J=\oint_C \frac{\log^2(z)}{z^2-1}\,dz$$

where $C$ is the classical keyhole contour with (i) the branch cut along the non-negative real axis and (ii) with deformations around $z=1$. Applying the residue theorem, it is easy to see that $J=i\pi^3$. Therefore, we find that $$\begin{align} J&=i\pi^3\\\\ &=\int_{0}^{\infty}\frac{\log^2(x)}{x^2-1}\,dx-\text{PV}\int_0^\infty \frac{\left(\log(x)+i2\pi\right)^2}{x^2-1}\,dx\\\\ &=-i4\pi\int_0^\infty \frac{\log(x)}{x^2-1}\,dx\\\\ &+\color{blue}{(4\pi^2)\text{PV}\left(\int_0^\infty \frac{1}{x^2-1}\,dx\right)}\\\\ &+\color{red}{(4\pi^2)\lim_{\epsilon \to 0^+}\int_{\pi}^{2\pi} \frac{1}{(1+\epsilon e^{i\phi})^2-1}\,(i\epsilon e^{i\phi})\,d\phi}\\\\ &=-i4\pi\int_0^\infty \frac{\log(x)}{x^2-1}\,dx+\color{blue}{0}+\color{red}{i2\pi^3}\tag 4 \end{align}$$

Finally, solving $(4)$ for the integral of interest yields

$$\frac12\int_0^\infty \frac{\log(x)}{x^2-1}\,dx=\frac{\pi^2}{8}$$

and hence we find that the series of interest is

$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\frac{\pi^2}{8}}$$

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METHODOLGY $2$

I thought it might be instructive to include another way forward that relies strictly on a well-known method of contour integration. The true potential usefulness of this posted solution is that it presents a methodology that can be applied to a wide class of problems for which some of the simpler ways forward to evaluate the specific series of interest fail. It is to that end that we proceed.

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Observe that the function $\pi \cot(\pi z)$ has simple poles at $z=n\in \mathbb{Z}$. Let $\displaystyle f(z)=\frac{\pi \cot(\pi z)}{(2z+1)^2}$, which has a second order pole at $z=-1/2$ and simple poles at $z=n$.

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Residues of $f(z)$:

The residues of $f(z)$ at $z=n$ are

$$\begin{align} \text{Res}(f(z),z=n)&=\lim_{z\to n}\frac{(z-n)\pi \cot(\pi z)}{(2z+1)^2}\\\\ &=\frac{1}{(2n+1)^2} \end{align}$$

while the residue of $f(z)$ at $z=-1/2$ is

$$\begin{align} \text{Res}(f(z),z=-1/2)&=\lim_{z\to -1/2}\frac{d}{dz}\left(\frac{(z+1/2)^2\pi \cot(\pi z)}{(2z+1)^2}\right)\\\\ &=-\frac{\pi^2}{4} \end{align}$$

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Contour Integration of $f(z)$:

We choose a contour $C$ to be the circle of radius $N+1/2$, centered at the origin. The Residue Theorem guarantees that

$$\oint_C f(z)\,dz=2\pi i \left(\sum_{n=-N}^N \frac{1}{(2n+1)^2}-\frac{\pi^2}{4}\right) \tag 1$$

In addition, as $N\to \infty$, the contour integral on the left-hand side of $(1)$ approaches zero.

Putting it all together, we see that

$$\begin{align} \sum_{-\infty}^\infty\frac{1}{(2n+1)^2}&=2\sum_{n=0}^\infty\frac{1}{(2n+1)^2}\\\\ &=\frac{\pi^2}{4} \end{align}$$

whereby we obtain the coveted result

$$\sum_{n=0}^\infty\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$

And we are done!

Answer 4

Not a geometric series.

Hint: consider Basel Problem. Consider sum over even terms. Subtract it.