Relationship between Cesàro's Lemma and Stolz–Cesàro theorem Probability with Martingales:





From Wiki:



What is their relationship? Does any imply the other?
 A: It looks like I can prove Stolz–Cesàro theorem from Cesàro's lemma.

What I tried:
Choose $\{v_k\}_{k=0}^{\infty}$ s.t.
$$v_k = \frac{a_k - a_{k-1}}{b_k - b_{k-1}}$$
where $\{a_k\}_{k=0}^{\infty}$ is any sequence such that
$$\left\{\frac{a_k - a_{k-1}}{b_k - b_{k-1}}\right\}_{k=0}^{\infty}$$
is convergent.
By Cesàro's Lemma, we have
$$\lim \frac{1}{b_n} \sum_{k=1}^{n} \frac{(b_k - b_{k-1})(a_k - a_{k-1})}{b_k - b_{k-1}} = v_{\infty}$$
$$\to v_{\infty} = \lim \frac{1}{b_n} \sum_{k=1}^{n} (a_k - a_{k-1})$$
$$\to v_{\infty} = \lim \frac{1}{b_n} (a_n - a_0)$$
$$\to v_{\infty} = \lim \frac{a_n}{b_n} - \lim \frac{a_0}{b_n}$$
$$\to v_{\infty} = \lim \frac{a_n}{b_n}$$
QED

How about proving the converse?
Choose $\{a_k\}_{k=0}^{\infty}$ s.t.
$$a_k = v_k(b_k - b_{k-1}) + a_{k-1}$$
where $\{v_k\}_{k=1}^{\infty}$ any convergent sequence.
Then
$$\left\{\frac{a_k - a_{k-1}}{b_k - b_{k-1}}\right\}_{k=0}^{\infty}$$
is convergent.
By Stolz–Cesàro theorem,
$$\lim \frac{a_{n+1} - a_n}{b_{n+1} - b_n} = \lim \frac{a_n}{b_n}$$
$$\to \lim v_n = \lim \frac{a_n}{b_n}$$
$$\to v_{\infty} = \lim \frac{a_n}{b_n}$$
Then
$$\lim \frac{1}{b_n} \sum_{k=1}^{n} (b_k - b_{k-1})v_k$$
$$= \lim \frac{1}{b_n} \sum_{k=1}^{n} (b_k - b_{k-1})\frac{a_k - a_{k-1}}{b_k - b_{k-1}}$$
$$= \lim \frac{1}{b_n} \sum_{k=1}^{n} (a_k - a_{k-1})$$
$$= \lim \frac{1}{b_n} (a_n - a_0)$$
$$= \lim \frac{1}{b_n} (a_n - a_0)$$
$$= \lim \frac{a_n}{b_n} - \lim \frac{a_0}{b_n}$$
$$= \lim \frac{a_n}{b_n} - 0$$
$$= v_{\infty} - 0 = v_{\infty}$$
QED
