1. 2014
Nov
20

## Another Mathematica bug

Math is hard.

Not for Barbie, but for Mathematica.

I ran into a weird Mathematica bug while trying to evaluate the sum

$$\sum_{k=1}^{\infty} \biggl[-\frac{\pi^2}{6} + \psi'(k + 1) + H_k^2\biggr]\frac{z^k}{k!}$$

Split this into three parts. The first one is the well-known expansion of the exponential function

$$-\frac{\pi^2}{6}\sum_{k=1}^{\infty} \frac{z^k}{k!} = -\frac{\pi^2}{6}(e^z - 1)$$

The second is not the well-known expansion of the exponential function.

$$\sum_{k=1}^{\infty} \psi'(k + 1)\frac{z^k}{k!} \neq \frac{\pi^2}{6}(e^z - 1)$$

Obviously not, in fact, since if two power series are equal, $$\sum_i a_n z^n = \sum_i b_n z^n$$, for an infinite number of points, each of their coefficients have to be equal: $$\forall n,\ a_n = b_n$$. (You can show this by taking the difference of the two sides and plugging in a bunch of different values of $$z$$.)

I guess Mathematica doesn’t know that.

In[1] = Sum[PolyGamma[1, k + 1] z^k/k!, {k, 1, Infinity}]
Out[1] = 1/6(-1 + E^z)Pi^2


I had my hopes up for …