How is the infinite sum of a series calculated by symbolic math? 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?
 A: 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.
A: 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 $x=-1$.
A: 
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!
