Relationship between Cesàro's Lemma and Stolz–Cesàro theorem

Question

Probability with Martingales:

---

---

---

From Wiki:

---

---

What is their relationship? Does any imply the other?

Answer 1

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