A generalization of the Stolz-Cesàro: For $k\ge 1$, $\lim_{n\to\infty }\frac{a_{n+k}-a_n}{b_{n+k}-b_n}=l$ implies $\lim_{n\to\infty}\frac{a_n}{b_n}=l$

Question

Stolz-Cesàro theorem case $\frac{*}{\infty}$:- If $b_n $ is a monotone increasing sequence and $\lim \limits_{n \to \infty} b_n = \infty $, and if $\lim \limits_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}- b_n}= l \in \overline{\mathbb{R}} $, then $\lim \limits_{n \to \infty} \frac{a_n }{b_n}=l$.

Stolz-Cesàro theorem case $\frac{0}{0}$:- If $b_n $ is a monotone decreasing sequence and $\lim \limits_{n \to \infty} b_n = \lim \limits_{n \to \infty} a_n = 0 $, and if $\lim \limits_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}- b_n}= l \in \overline{\mathbb{R}} $, then $\lim \limits_{n \to \infty} \frac{a_n }{b_n}=l$.

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I was solving this limit :If $a_n \to a \in \mathbb{R}$ find $$\lim\limits_{n \to \infty } \frac{\sum\limits_{k=1}^n (-1)^{n-k} 2^k a^k}{2^n}$$

The numerator is a bit annoying to apply Stolz-Cesàro theorem to because of that $(-1)^{n-k}$.

The numerator is $\sum\limits_{k=1}^n (-1)^{n-k} 2^k a^k$ when $n$ is odd and $-\sum\limits_{k=1}^n (-1)^{n-k} 2^k a^k$ when $n$ is even.

It would have been much easier if $\lim\limits_{n \to \infty } \frac{a_{n +2}- a_n}{b_{2+n}- b_n}= l$ was true, and out of curiosity, I tried to use this statement here:

$$\lim \limits_{n \to \infty} \frac{2^{n+2}a_{n+2}-2^{n+1}a_{n+1}}{2^{n+2}-2^{n}}=\lim \limits_{n \to \infty} \frac{4a_{n+2}-2a_{n+1}}{3} =\frac{2a}{3}$$

Which, in fact, is the correct limit!

This got me wondering if there is a generalisation of the Stolz-Cesàro Theorem.

Generalized Stolz-Cesàro theorem case $\frac{*}{\infty}$:- If $b_n $ is a monotone increasing sequence and $\lim \limits_{n \to \infty} b_n = \infty $, and $\exists k \in \mathbb{N}$ st $\lim \limits_{n \to \infty} \frac{a_{n+k}-a_n}{b_{n+k}- b_n}= l \in \overline{\mathbb{R}} $, then $\lim \limits_{n \to \infty} \frac{a_n }{b_n}=l$.

Generalized Stolz-Cesàro theorem case $\frac{0}{0}$:- If $b_n $ is a monotone decreasing sequence and $\lim \limits_{n \to \infty} b_n = \lim \limits_{n \to \infty} a_n = 0 $, and if $\exists k \in \mathbb{N}$ st $\lim \limits_{n \to \infty} \frac{a_{n+k}-a_n}{b_{n+k}- b_n}= l \in \overline{\mathbb{R}} $, then $\lim \limits_{n \to \infty} \frac{a_n }{b_n}=l$.

Example: $\lim_limits_{n \to \infty}\frac{(-1)^n}{n } =0$ but the ordinal Stolz-Cesàro Limit don't exist , if the generalisation was true then $\lim \limits_{n \to \infty} \frac{a_{n+2} -a_{n}}{b_{n+2}-b_{n}}= 0$ (I know this is a trivial example )

If it is not true, then what are the sufficient conditions for $a_n​,b_n$​ so that this generalisation of the Stolz-Cesàro theorem becomes true?

Answer 1

Let $(b_n)$ be an increasing sequence tending to $\infty$, and let $(a_n)$ be another sequence. Suppose $k\ge1$ is an integer such that $\lim_{n\to\infty} \dfrac{a_{n+k}-a_n}{b_{n+k}-b_n} = L$.

For every integer $j\in\{0,\dots,k-1\}$, define $a^{(j)}_m = a_{km+j}$ and $b^{(j)}_m = b_{km+j}$. Note that $(b^{(j)}_m)$ is increasing and tends to $\infty$ for each $j$. Furthermore, $$ \lim_{m\to\infty} \frac{a^{(j)}_{m+1}-a^{(j)}_m}{b^{(j)}_{m+1}-b^{(j)}_m} = \lim_{m\to\infty} \frac{a_{km+j+k}-a_{km+j}}{b_{km+j+k}-b_{km+j}} = \lim_{n\to\infty} \frac{a_{n+k}-a_n}{b_{n+k}-b_n} = L $$ by assumption, where the middle equality holds because $(a_{km+j})$ is a subsequence of $(a_n)$. By the original Stolz–Cesàro theorem, we conclude that $$ \lim_{m\to\infty} \frac{a_{km+j}}{b_{km+j}} = \lim_{m\to\infty} \frac{a^{(j)}_m}{b^{(j)}_m} = L $$ for every $j\in\{0,\dots,k-1\}$.

But now $\bigl( \frac{a_n}{b_n} \bigr)$ is the union of the finitely many subsequences $\bigl( \frac{a_{km+j}}{b_{km+j}} \bigr)$, each of which converges to $L$; we conclude that the entire sequence $\bigl( \frac{a_n}{b_n} \bigr)$ converges to $L$.