# 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 $\lim_{n\to\infty}\frac{a_n}{b_n}=l$

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 $$\lim \limits_{n \to \infty} \frac{a_n }{b_n}=l$$.

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?

• Your original limit is trivial and there is not need for fancy theorems: $$\frac{\sum\limits_{k=1}^n (-1)^{n-k} 2^k a^k}{2^n}=(-2)^{-n}\sum^n_{k=1}(-2a)^k=(-2)^{-n}\frac{(-2a)^{n+1}-1}{1+2a}$$ Commented Dec 9, 2023 at 1:56
• @Mittens I know that but this Limit is what gave me the idea of this "fancy theorem"
– pie
Commented Dec 9, 2023 at 1:57
• I'm not trying to discourage you. It isn just that in your expression, if $|2a|<1$, then the limit you are after is $0$, the factor $(-1)^{n-k}$ is not hard to control . Commented Dec 9, 2023 at 3:22
• Doesn't the proposed generalization follow immediately from the original version upon separately considering the subsequences $\{a_{km+r}\}_{m=1}^\infty$ and $\{b_{km+r}\}_{m=1}^\infty$ (for $r=0,\dots,k-1$)? Commented Dec 9, 2023 at 8:40
• @GregMartin I couldn't prove that. If you can elaborate more I will appreciate it
– pie
Commented Dec 9, 2023 at 15:47

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 $$L$$.