Partial Progression?

The infinite summation of AP is Cesàro summable if the following limit $\mathscr L$ exists:

$\begin{aligned} \mathscr L & = \lim_{n \to \infty} \frac 1n \sum_{k=1}^n \sum_{i=1}^k a_i \\ & = \lim_{n \to \infty} \frac 1n \sum_{k=1}^n (Ak + D(k-1)) \\ & = \lim_{n \to \infty} \frac 1n \left(\frac {n(n+1)}2 (A+D)-nD \right) \\ & = \lim_{n \to \infty} \left(\frac {n+1}2 (A+D)-D \right) \\ & = \infty \end{aligned}$

$\boxed{\text{No, it is not Cesàro summable.}}$

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In response to @Pi Han Goh , thanks for letting me know about Stolz–Cesàro theorem .

Yes, it can be solved using Stolz–Cesàro theorem 1 (the $\infty / \infty$ case).

We are checking if $\displaystyle \lim_{n \to \infty} \dfrac {\sum_{k=1}^n s_k}n$ exists. Putting $a_n = \sum_{k=1}^n s_k$ and $b_n = n$ , then if $\displaystyle \lim_{n \to \infty} \frac {a_{n+1}-a_n}{b_{n+1}-b_n} = l$ exists, then $\displaystyle \lim_{n \to \infty} \frac {a_n}{b_n} = l$ .

$\begin{aligned} l & = \lim_{n \to \infty} \frac {a_{n+1}-a_n}{b_{n+1}-b_n} \\ & = \lim_{n \to \infty} \frac {\sum_{k=1}^{n+1} s_k - \sum_{k=1}^n s_k}{(n+1)-n} \\ & = \lim_{n \to \infty} \frac{s_{n+1}}1 \\ & = \lim_{n \to \infty} ((n+1)A + nD) \\ & = \infty \end{aligned}$

$\implies \displaystyle \lim_{n \to \infty} \dfrac {\sum_{k=1}^n s_k}n$ does not exist.

$\boxed{\text{No, it is not Cesàro summable.}}$