In this post: Cesaro summable implies Abel summable, in the proof that Cesaro summable implies Abel summable, the following relation is used:
$$\sum_{n=0}^\infty s^1_n x^n = (1-x)^{-1} \sum_{n=0}^\infty s^0_n x^n = (1-x)^{-2} \sum_{n=0}^\infty a_n x^n$$
I know that the series $\sum_{n = 0}^{\infty} ar^n = \frac{a}{1-r}$ for $|r| \lt 1$, so the standard geometric series. What I see in the relation is the following: $a = s_n^1, \ r = x$ and $s_n^1 = \sum_{n = 0}^{\infty} s_n^0$ (as defned in the proof).
Following this, I would expect $$\sum_{n=0}^\infty s^1_n x^n = \frac{s_n^1}{1-x} = (1-x)^{-1} \sum_{n = 0}^{\infty}s_n^0$$.
But why is it $\sum_{n = 0}^{\infty}s_n^0 x^n$? Where does the $x^n$ come from?