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?

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.
