For $0<x<1$, the Abel transform of a series is $(1-x)^2\sum_{n\ge0}(n+1)\sigma_nx^n$, where $\sigma_n$ are the Cesaro means of its partial sums

Given: The coefficients, partial sums, and Cesaro means in the statement.

The canonical natural $\iota(n+1)$ is positive. Thus, putting $T_n:=\sum_{k=0}^{n}S_k$, the definition of $\sigma_n$ gives $T_n=\iota(n+1)\sigma_n$. Also $S_n=T_n-T_{n-1}$ with $T_{-1}:=0$; putting $S_{-1}:=0$ gives $a_n=S_n-S_{n-1}$ for every $n\ge0$ (Abel summability by $\lim_{x\uparrow1}\sum a_nx^n$ and Cesaro summability by the Cesaro means of the partial sums, Finite sums and finite products, by recursion, The canonical natural $\iota(n) = n \cdot 1_F$ of a field, Canonical naturals are positive and strictly increasing).

For $0<x<1$, the geometric series is absolutely convergent because its terms are nonnegative. Its Cauchy product with itself has coefficient $\sum_{k=0}^{n}x^kx^{n-k}=\iota(n+1)x^n$ and sum $1/(1-x)^2$ (For $|r| < 1$, $\sum_{k \ge 0} r^k = 1/(1-r)$, and for $|r| \ge 1$ the series diverges, Monotonicity of $x \mapsto x^n$ and of $n \mapsto a^n$, Basic properties of the absolute value, The Cauchy product of two series: $c_n = \sum_{k=0}^{n} a_k b_{n-k}$, If $\sum a_k$ and $\sum b_k$ both converge absolutely then their Cauchy product converges absolutely, with sum $AB$, Finite sums and finite products, by recursion, The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

Convergent real series may be added, subtracted and scaled term by term (Convergent series add and scale termwise).

A nonnegative series dominated termwise by a convergent nonnegative series converges (If $0 \le a_k \le b_k$ eventually, convergence of $\sum b_k$ gives convergence of $\sum a_k$, and divergence of $\sum a_k$ gives divergence of $\sum b_k$).