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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?

syphracos
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  • There is a minor typo in the linked answer, it's $$s_n^1 := \sum_{k=0}^{n} s_k^0$$ and so it depends on $n$, so you cannot get it out of the sum. – Bruno B Nov 28 '22 at 12:42
  • @BrunoB so if I understand correctly, you HAVE to carry the $x^n$ to the next sum? – syphracos Nov 28 '22 at 12:50

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