Cesaro summable implies Abel summable I've looked through the Stack old questions, and searched the net, and I haven't found a proof that Cesaro summability implies Abel summability. Is the proof extremely complicated? Does anyone know a good reference?
I am also interested in the proof that if $\sum a_n$ and $\sum b_n$ are convergent, then the Cauchy product of the two sequences is Cesaro summable.
 A: Proof that Cesaro summable implies Abel summable : let $(a_n)_n$ be a sequence. Denote
$$s^0_n = \sum_{n=0}^n a_n ~, \quad s^1_n = \sum_{n=0}^n s^0_n.$$
Suppose hat $\sum a_n$ is Cesaro summable, i.e. $\lim_n s^1_n/(n+1) = L$.
Note that (easy calculation)
$$\sum_{n=0}^\infty s^1_n x^n = (1-x)^{-1} \sum_{n=0}^\infty s^0_n x^n = (1-x)^{-2} \sum_{n=0}^\infty a_n x^n$$
(this calculation also prove that all the series converge for $|x|<1$).
We have to prove that $\lim_{x \to 1} (1-x)^2 \sum_n s^1_n x^n = L$. Note that $\sum (n+1) x^n = (1-x)^{-2}$. Let  $N>0$ and $x \in [0,1)$ and write
$$(1-x)^2\sum_{n=0}^\infty s^1_n x^n = \sum_{n=0}^\infty (n+1)(1-x)^2x^n \times \frac{s^1_n}{n+1},$$
$$L = \sum_{n=0}^\infty (n+1)(1-x)^2x^n \times L.$$
Hence
$$\left|(1-x)^2 \sum_{n=0}^\infty s^1_n x^n - L \right|
\leq \sum_{n=0}^N (n+1)(1-x)^2x^n \times |\tfrac{s^1_n}{n+1}-L| + \sum_{n=N+1}^\infty (n+1)(1-x)^2x^n \times |\tfrac{s^1_n}{n+1}-L|.$$
From this formula is easy to conclude (the second term in the RHS is small for $N$ big indepandently from $x$, and the first term goes to $0$ for $N$ fixed and $x\to 1$)
Proof that $\sum (a*b)_n$ is Cesaro summable. Let $\sum a_n$ $\sum_n b_n$ be two series and let : $c_n = \sum_{p+q=n} a_p b_q$ and $X_N = \sum_{n=0}^N x_n$ (where $x=a,b,c$). I let you show that
$$ \sum_{N=0}^M C_N = \sum_{P+Q=M} A_P B_Q.$$
(Hint : you can show for example that both sides are equal to $\sum_{p+q \leq M} (M+1-p-q)a_p b_q$). From this, it is easy to conclude that $\lim_M (M+1)^{-1} \sum_0^M C_N =A.B$ (where $A=\sum a_n = \lim_n A_n$, $B=\sum b_n = \lim_n B_n$), because if $M$ is big then $P$ and $Q$ are big hence, $A_P$ is closed to $A$ and $B_Q$ is closed to $B$. 
