Euler transform

Remarks

Example: for $a_n = \frac{1}{n+1}$ one has $(-1)^n \Delta^n a_0 = \frac{1}{n+1}$, so the alternating harmonic series becomes $\sum_{n \ge 0} \frac{1}{(n+1) 2^{n+1}}$, which equals $\ln 2$. For $a_n = 1$ the series $1 - 1 + 1 - \cdots$ diverges, but $\Delta^n a_0 = 0$ for $n \ge 1$ and the transform is $\frac12$; this is how Euler summation assigns the value $\frac12$ to it. Non-example: for $a_n = 3^n$ one gets $\Delta^n a_0 = 2^n$, and the transform $\sum_{n \ge 0} \frac{(-1)^n}{2}$ diverges, so the transform does not always produce a convergent series.