How is the infinite sum of a series calculated by symbolic math?

Question

I wonder how Wolfram can solve this series and provide the solution symbolically:

$$\sum_{k=1}^\infty\frac 1{(2k-1)^4}$$

In this particular case I know how to use a Fourier series on a triangle function to get the result by employing Parseval's theorem, but this is only a particular example. The proof for $\sum_{k=1}^{k=\infty} \frac{1}{k^2} = \frac{\pi^2}{6}$ was found by Euler and uses a taylor series of a special function.

But is there a recipe working correctly for each possible series? I cannot imagine that such algorithm exists. But how can Wolfram do it?

Answer 1

I bet that WolframAlpha has a database of the most common expressions. Wolfram publicly provides a huge list of functions, containing various series expansions of most functions, so it's not very hard to build an optimized database with the mapping series$\to$function.

For example, WolframAlpha returns $$\sum_{k=1}^\infty\frac 1{(k+a)^n} = \zeta(n, a+1)\text.$$

To evaluate your input, apply a general technique: try to eliminate any integer factors from the summation variable. $$\sum_{k=1}^\infty\frac 1{(2k-1)^4}=\sum_{k=1}^\infty\frac1{2^4}\cdot\frac 1{(k-\frac12)^4}=\frac1{16}\cdot\sum_{k=1}^\infty\frac 1{(k-\frac12)^4}=\frac1{16}\cdot\zeta(4, 1/2)$$

Now, without even knowing what this $\zeta$ function is, we just need to know the value at the point $(4,1/2)$, and we're done.

There is a database of special values of this function. It does not contain $(4, 1/2)$ though.

So, try the transformation database. It has $$\zeta(n, 1/2)=(2^n-1)\zeta(n)$$ which brings us to another $\zeta$ function (one argument). Again, we need to know nothing about the function, just its value at $n=4$. Here, the database yields $$\zeta(4)=\frac{\pi^4}{90}$$

and we're done.

---

Disclaimer: I don't know exactly how WolframAlpha works, so this is just a guess.

Answer 2

If you can show that

$$\sum_{n=1}^\infty\frac1{n^2+x}=-\frac1{2x}+\frac\pi{2\sqrt x\tanh(\pi\sqrt x)}$$

Then just differentiate a few times and plug in appropriate values for $x$. You can likewise solve the following series:

$$\sum_{n=1}^\infty\frac1{(ak+x)^2}$$

Differentiate twice and put $a=2$ and $x=-1$.

Answer 3

I do not know how symbolic programs evaluate series in closed form. But I thought it might be instructive to see one methodology to solve the problem of interest. It is to that end we now proceed.

---

Note that $\zeta(4)=\sum_{n=1}^\infty\frac{1}{n^4}$. Next, we can write $\zeta(4)$ as

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

by pairing even and odd terms. But $\sum_{n=1}^\infty\frac{1}{(2n)^4}=\frac1{16}\zeta(4)$. Using this in $(1)$ and solving for the term of interest reveals

$$\sum_{n=1}^\infty\frac{1}{(2n-1)^4}=\frac{15}{16}\zeta(4)$$

---

FINDING $\displaystyle \zeta(4)$:

To find $\zeta(4)$, we can use contour integration. Let $f(z)=\frac{\pi \cot(\pi z)}{z^4}$ and $C$ be a circular contour of radius $N+1/2$. Then, we have

$$\begin{align} \oint_C \frac{\pi \cot(\pi z)}{z^4}\,dz&=2\pi i \sum_{n=-N}^N \text{Res}\left(\frac{\pi \cot(\pi z)}{z^4}, z=n\right)\\\\ &=2\pi i \sum_{|n|\ge 1} \frac1{n^4}+2\pi i \text{Res}\left(\frac{\pi \cot(\pi z)}{z^4}, z=0\right) \tag 2 \end{align}$$

As $N\to \infty$, the integral in $(2)$ approaches $0$. Hence, from $(2) we have

$$\sum_{n=1}^\infty \frac1{n^4}=-\frac12 \text{Res}\left(\frac{\pi \cot(\pi z)}{z^4}, z=0\right) \tag 3$$

---

EVALUATING THE RESIDUE AT $0$:

Noting that $z=0$ is a fifth-order pole, we could calculate the residue by applying the formula

$$\text{Res}\left(\frac{\pi \cot(\pi z)}{z^4}, z=0\right)=\frac1{4!}\lim_{z\to 0}\frac{d^4 (\pi z \cot(\pi z))}{dz^4} $$

Rather than pursue this approach, we expand the integrand as

$$\begin{align} \frac{\pi\cot(\pi z)}{z^4}&= \frac{1-\frac12(\pi z)^2+\frac{1}{24}(\pi z)^4+O(z^6)}{ z^5\left(1-\frac16 (\pi z)^2+\frac{1}{120}(\pi z)^4+O(z^6)\right)}\\\\ &=\frac{1}{z^5}\left(1-\frac12(\pi z)^2+\frac{1}{24}(\pi z)^4+O(z^6)\right)\left(1+\frac16 (\pi z)^2+\frac{7}{360}(\pi z)^4+O(z^6)\right) \end{align}$$

The residue is the coefficient on the term $z^{-1}$. Therefore,

$$\text{Res}\left(\frac{\pi \cot(\pi z)}{z^4}, z=0\right)=\left(-\frac1{12}+\frac{1}{24}+\frac{1}{360}\right)\,\pi^4=-\frac{\pi^4}{45} \tag 4$$

---

Using $(4)$ in $(3)$ we find that $\zeta(4)=\frac{\pi^4}{90}$. And finally, we have

$$\sum_{n=1}^\infty\frac{1}{(2n-1)^4}=\frac{\pi^4}{96}$$

And we are done!