Stolz–Cesàro theorem

In mathematics, the Stolz–Cesàro theorem is a criterion for proving the convergence of a sequence. The theorem is named after mathematicians Otto Stolz and Ernesto Cesàro, who stated and proved it for the first time.

The Stolz–Cesàro theorem can be viewed as a generalization of the Cesàro mean, but also as a l'Hôpital's rule for sequences.

Statement of the theorem for the ∙/∞ case

Let ${\displaystyle (a_{n})_{n\geq 1}}$ and ${\displaystyle (b_{n})_{n\geq 1}}$ be two sequences of real numbers. Assume that ${\displaystyle (b_{n})_{n\geq 1}}$ is a strictly monotone and divergent sequence (i.e. strictly increasing and approaching ${\displaystyle +\infty }$, or strictly decreasing and approaching ${\displaystyle -\infty }$) and the following limit exists:

${\displaystyle \lim _{n\to \infty }{\frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}=l.\ }$

Then, the limit

${\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=l.\ }$

Statement of the theorem for the 0/0 case

Let ${\displaystyle (a_{n})_{n\geq 1}}$ and ${\displaystyle (b_{n})_{n\geq 1}}$ be two sequences of real numbers. Assume now that ${\displaystyle (a_{n})\to 0}$ and ${\displaystyle (b_{n})\to 0}$ while ${\displaystyle (b_{n})_{n\geq 1}}$ is strictly monotone. If

${\displaystyle \lim _{n\to \infty }{\frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}=l,\ }$

then

${\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=l.\ }$[1]

Proofs

Proof of the theorem for the ${\displaystyle \cdot /\infty }$ case

Case 1: suppose ${\displaystyle (b_{n})}$ strictly increasing and divergent to ${\displaystyle +\infty }$, and ${\displaystyle l<\infty }$. By hypothesis, we have that for all ${\displaystyle \epsilon /2>0}$ there exists ${\displaystyle \nu >0}$ such that ${\displaystyle \forall n>\nu }$

${\displaystyle \left|\,{\frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}-l\,\right|<{\frac {\epsilon }{2}},}$

which is to say

${\displaystyle l-\epsilon /2<{\frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}<l+\epsilon /2,\quad \forall n>\nu .}$

Since ${\displaystyle (b_{n})}$ is strictly increasing, ${\displaystyle b_{n+1}-b_{n}>0}$, and the following holds

${\displaystyle (l-\epsilon /2)(b_{n+1}-b_{n})<a_{n+1}-a_{n}<(l+\epsilon /2)(b_{n+1}-b_{n}),\quad \forall n>\nu }$.

Next we notice that

${\displaystyle a_{n}=[(a_{n}-a_{n-1})+\dots +(a_{\nu +2}-a_{\nu +1})]+a_{\nu +1}}$

thus, by applying the above inequality to each of the terms in the square brackets, we obtain

${\displaystyle {\begin{aligned}&(l-\epsilon /2)(b_{n}-b_{\nu +1})+a_{\nu +1}=(l-\epsilon /2)[(b_{n}-b_{n-1})+\dots +(b_{\nu +2}-b_{\nu +1})]+a_{\nu +1}<a_{n}\\&a_{n}<(l+\epsilon /2)[(b_{n}-b_{n-1})+\dots +(b_{\nu +2}-b_{\nu +1})]+a_{\nu +1}=(l+\epsilon /2)(b_{n}-b_{\nu +1})+a_{\nu +1}.\end{aligned}}}$

Now, since ${\displaystyle b_{n}\to +\infty }$ as ${\displaystyle n\to \infty }$, there is an ${\displaystyle n_{0}>0}$ such that ${\displaystyle b_{n}\gneq 0}$ for all ${\displaystyle n>n_{0}}$, and we can divide the two inequalities by ${\displaystyle b_{n}}$ for all ${\displaystyle n>\max\{\nu ,n_{0}\}}$

${\displaystyle (l-\epsilon /2)+{\frac {a_{\nu +1}-b_{\nu +1}(l-\epsilon /2)}{b_{n}}}<{\frac {a_{n}}{b_{n}}}<(l+\epsilon /2)+{\frac {a_{\nu +1}-b_{\nu +1}(l+\epsilon /2)}{b_{n}}}.}$

The two sequences (which are only defined for ${\displaystyle n>n_{0}}$ as there could be an ${\displaystyle N\leq n_{0}}$ such that ${\displaystyle b_{N}=0}$)

${\displaystyle c_{n}^{\pm }:={\frac {a_{\nu +1}-b_{\nu +1}(l\pm \epsilon /2)}{b_{n}}}}$

are infinitesimal since ${\displaystyle b_{n}\to +\infty }$ and the numerator is a constant number, hence for all ${\displaystyle \epsilon /2>0}$ there exist ${\displaystyle n_{\pm }>n_{0}>0}$, such that

${\displaystyle {\begin{aligned}&|c_{n}^{+}|<\epsilon /2,\quad \forall n>n_{+},\\&|c_{n}^{-}|<\epsilon /2,\quad \forall n>n_{-},\end{aligned}}}$

therefore

${\displaystyle l-\epsilon <l-\epsilon /2+c_{n}^{-}<{\frac {a_{n}}{b_{n}}}<l+\epsilon /2+c_{n}^{+}<l+\epsilon ,\quad \forall n>\max \lbrace \nu ,n_{\pm }\rbrace =:N>0,}$

which concludes the proof. The case with ${\displaystyle (b_{n})}$ strictly decreasing and divergent to ${\displaystyle -\infty }$, and ${\displaystyle l<\infty }$ is similar.

Case 2: we assume ${\displaystyle (b_{n})}$ strictly increasing and divergent to ${\displaystyle +\infty }$, and ${\displaystyle l=+\infty }$. Proceeding as before, for all ${\displaystyle {\frac {3}{2}}M>0}$ there exists ${\displaystyle \nu >0}$ such that for all ${\displaystyle n>\nu }$

${\displaystyle {\frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}>{\frac {3}{2}}M.}$

Again, by applying the above inequality to each of the terms inside the square brackets we obtain

${\displaystyle a_{n}>{\frac {3}{2}}M(b_{n}-b_{\nu +1})+a_{\nu +1},\quad \forall n>\nu ,}$

and

${\displaystyle {\frac {a_{n}}{b_{n}}}>{\frac {3}{2}}M+{\frac {a_{\nu +1}-{\frac {3}{2}}Mb_{\nu +1}}{b_{n}}},\quad \forall n>\max\{\nu ,n_{0}\}.}$

The sequence ${\displaystyle (c_{n})_{n>n_{0}}}$ defined by

${\displaystyle c_{n}:={\frac {a_{\nu +1}-{\frac {3}{2}}Mb_{\nu +1}}{b_{n}}}}$

is infinitesimal, thus

${\displaystyle \forall M/2>0\,\exists {\bar {n}}>n_{0}>0{\text{ such that }}-M/2<c_{n}<M/2,\,\forall n>{\bar {n}},}$

combining this inequality with the previous one we conclude

${\displaystyle {\frac {a_{n}}{b_{n}}}>{\frac {3}{2}}M+c_{n}>M,\quad \forall n>\max\{\nu ,{\bar {n}}\}=:N.}$

The proofs of the other cases with ${\displaystyle (b_{n})}$ strictly increasing or decreasing and approaching ${\displaystyle +\infty }$ or ${\displaystyle -\infty }$ respectively and ${\displaystyle l=\pm \infty }$ all proceed in this same way.

Proof of the theorem for the ${\displaystyle 0/0}$ case

Case 1: we first consider the case with ${\displaystyle l<\infty }$ and ${\displaystyle (b_{n})}$ strictly increasing. This time, for each ${\displaystyle m>0}$, we can write

${\displaystyle a_{n}=(a_{n}-a_{n-1})+\dots +(a_{m+\nu +1}-a_{m+\nu })+a_{m+\nu },}$

and

${\displaystyle {\begin{aligned}&(l-\epsilon /2)(b_{n}-b_{\nu +m})+a_{\nu +m}=(l-\epsilon /2)[(b_{n}-b_{n-1})+\dots +(b_{\nu +m+1}-b_{\nu +m})]+a_{\nu +m}<a_{n}\\&a_{n}<(l+\epsilon /2)[(b_{n}-b_{n-1})+\dots +(b_{\nu +m+1}-b_{\nu +m})]+a_{\nu +m}=(l+\epsilon /2)(b_{n}-b_{\nu +m})+a_{\nu +m}.\end{aligned}}}$

The two sequences

${\displaystyle c_{m}^{\pm }:={\frac {a_{\nu +m}-b_{\nu +m}(l\pm \epsilon /2)}{b_{n}}}}$

are infinitesimal since by hypotehsis ${\displaystyle a_{m},b_{m}\to 0}$, thus for all ${\displaystyle \epsilon /2>0}$ there are ${\displaystyle n^{\pm }>0}$ such that

${\displaystyle {\begin{aligned}&|c_{m}^{+}|<\epsilon /2,\quad \forall m>n_{+},\\&|c_{m}^{-}|<\epsilon /2,\quad \forall m>n_{-},\end{aligned}}}$

thus, choosing ${\displaystyle m}$ appropriately (which is to say, taking the limit with respect to ${\displaystyle m}$) we obtain

${\displaystyle l-\epsilon <l-\epsilon /2+c_{m}^{-}<{\frac {a_{n}}{b_{n}}}<l+\epsilon /2+c_{m}^{+}<l+\epsilon ,\quad \forall n>\max\{\nu ,n_{0}\},}$

which concludes the proof.

Case 2: we assume ${\displaystyle l=+\infty }$ and ${\displaystyle (b_{n})}$ strictly increasing. For all ${\displaystyle {\frac {3}{2}}M>0}$ there exists ${\displaystyle \nu >0}$ such that for all ${\displaystyle n>\nu }$

${\displaystyle {\frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}>{\frac {3}{2}}M.}$

Therefore, for each ${\displaystyle m>0}$

${\displaystyle {\frac {a_{n}}{b_{n}}}>{\frac {3}{2}}M+{\frac {a_{\nu +m}-{\frac {3}{2}}Mb_{\nu +m}}{b_{n}}},\quad \forall n>\max\{\nu ,n_{0}\}.}$

The sequence

${\displaystyle c_{m}:={\frac {a_{\nu +m}-{\frac {3}{2}}Mb_{\nu +m}}{b_{n}}}}$

converges to ${\displaystyle 0}$ (keeping ${\displaystyle n}$ fixed), hence

${\displaystyle \forall M/2>0\,\exists {\bar {n}}>0{\text{ such that }}-M/2<c_{m}<M/2,\,\forall m>{\bar {n}},}$

and, choosing ${\displaystyle m}$ conveniently, we conclude the proof

${\displaystyle {\frac {a_{n}}{b_{n}}}>{\frac {3}{2}}M+c_{m}>M,\quad \forall n>\max\{\nu ,n_{0}\}.}$

Applications and examples

The theorem concerning the ${\displaystyle \cdot /\infty }$ case has a few notable consequences which are useful in the computation of limits.

Arithmetic mean

Let ${\displaystyle (x_{n})}$ be a sequence of real numbers which converges to ${\displaystyle l}$, define

${\displaystyle a_{n}:=\sum _{m=1}^{n}x_{m}=x_{1}+\dots +x_{n},\quad b_{n}:=n}$

then ${\displaystyle (b_{n})}$ is strictly increasing and diverges to ${\displaystyle +\infty }$. We compute

${\displaystyle \lim _{n\to \infty }{\frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}=\lim _{n\to \infty }x_{n+1}=\lim _{n\to \infty }x_{n}=l}$

therefore

${\displaystyle \lim _{n\to \infty }{\frac {x_{1}+\dots +x_{n}}{n}}=\lim _{n\to \infty }x_{n}.}$

Given any sequence ${\displaystyle (x_{n})_{n\geq 1}}$ of real numbers, suppose that

${\displaystyle \lim _{n\to \infty }x_{n}}$

exists (finite or infinite), then

${\displaystyle \lim _{n\to \infty }{\frac {x_{1}+\dots +x_{n}}{n}}=\lim _{n\to \infty }x_{n}.}$

Geometric mean

Let ${\displaystyle (x_{n})}$ be a sequence of positive real numbers converging to ${\displaystyle l}$ and define

${\displaystyle a_{n}:=\log(x_{1}\cdots x_{n}),\quad b_{n}:=n,}$

again we compute

${\displaystyle \lim _{n\to \infty }{\frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}=\lim _{n\to \infty }\log {\Big (}{\frac {x_{1}\cdots x_{n+1}}{x_{1}\cdots x_{n}}}{\Big )}=\lim _{n\to \infty }\log(x_{n+1})=\lim _{n\to \infty }\log(x_{n})=\log(l),}$

where we used the fact that the logarithm is continuous. Thus

${\displaystyle \lim _{n\to \infty }{\frac {\log(x_{1}\cdots x_{n})}{n}}=\lim _{n\to \infty }\log {\Big (}(x_{1}\cdots x_{n})^{\frac {1}{n}}{\Big )}=\log(l),}$

since the logarithm is both continuous and injective we can conclude that

${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{x_{1}\cdots x_{n}}}=\lim _{n\to \infty }x_{n}}$.

Given any sequence ${\displaystyle (x_{n})_{n\geq 1}}$ of (strictly) positive real numbers, suppose that

${\displaystyle \lim _{n\to \infty }x_{n}}$

exists (finite or infinite), then

${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{x_{1}\cdots x_{n}}}=\lim _{n\to \infty }x_{n}.}$

Suppose we are given a sequence ${\displaystyle (y_{n})_{n\geq 1}}$ and we are asked to compute

${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{y_{n}}},}$

defining ${\displaystyle y_{0}=1}$ and ${\displaystyle x_{n}=y_{n}/y_{n-1}}$ we obtain

${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{x_{1}\dots x_{n}}}=\lim _{n\to \infty }{\sqrt[{n}]{\frac {y_{1}\dots y_{n}}{y_{0}\cdot y_{1}\dots y_{n-1}}}}=\lim _{n\to \infty }{\sqrt[{n}]{y_{n}}},}$

if we apply the property above

${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{y_{n}}}=\lim _{n\to \infty }x_{n}=\lim _{n\to \infty }{\frac {y_{n}}{y_{n-1}}}.}$

This last form is usually the most useful to compute limits

Given any sequence ${\displaystyle (y_{n})_{n\geq 1}}$ of (strictly) positive real numbers, suppose that

${\displaystyle \lim _{n\to \infty }{\frac {y_{n+1}}{y_{n}}}}$

exists (finite or infinite), then

${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{y_{n}}}=\lim _{n\to \infty }{\frac {y_{n+1}}{y_{n}}}.}$

Example 1

${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{n}}=\lim _{n\to \infty }{\frac {n+1}{n}}=1.}$

Example 2

${\displaystyle {\begin{aligned}\lim _{n\to \infty }{\frac {\sqrt[{n}]{n!}}{n}}&=\lim _{n\to \infty }{\frac {(n+1)!(n^{n})}{n!(n+1)^{n+1}}}\\&=\lim _{n\to \infty }{\frac {n^{n}}{(n+1)^{n}}}=\lim _{n\to \infty }{\frac {1}{(1+{\frac {1}{n}})^{n}}}={\frac {1}{e}}.\end{aligned}}}$

we used the representation of ${\displaystyle e}$ as the limit of a sequence in the last step.

Example 3

${\displaystyle \lim _{n\to \infty }{\frac {\log(n!)}{n\log(n)}}=\lim _{n\to \infty }{\frac {\log({\sqrt[{n}]{n!}})}{\log(n)}},}$

notice that

${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{n!}}=\lim _{n\to \infty }{\frac {(n+1)!}{n!}}=\lim _{n\to \infty }(n+1)}$

therefore

${\displaystyle \lim _{n\to \infty }{\frac {\log(n!)}{n\log(n)}}=\lim _{n\to \infty }{\frac {\log(n+1)}{\log(n)}}=1.}$

Example 4

Consider the sequence

${\displaystyle a_{n}=(-1)^{n}{\frac {n!}{n^{n}}}}$

this can be written as

${\displaystyle a_{n}=b_{n}\cdot c_{n},\quad b_{n}:=(-1)^{n},c_{n}:={\Big (}{\frac {\sqrt[{n}]{n!}}{n}}{\Big )}^{n},}$

the sequence ${\displaystyle (b_{n})}$ is bounded (and oscillating), while

${\displaystyle \lim _{n\to \infty }{\Big (}{\frac {\sqrt[{n}]{n!}}{n}}{\Big )}^{n}=\lim _{n\to \infty }(1/e)^{n}=0,}$

by the well-known limit, because ${\displaystyle 1/e<1}$; therefore

${\displaystyle \lim _{n\to \infty }(-1)^{n}{\frac {n!}{n^{n}}}=0.}$

History

The ∞/∞ case is stated and proved on pages 173—175 of Stolz's 1885 book and also on page 54 of Cesàro's 1888 article.

It appears as Problem 70 in Pólya and Szegő (1925).

The general form

Statement

The general form of the Stolz–Cesàro theorem is the following:[2] If ${\displaystyle (a_{n})_{n\geq 1}}$ and ${\displaystyle (b_{n})_{n\geq 1}}$ are two sequences such that ${\displaystyle (b_{n})_{n\geq 1}}$ is monotone and unbounded, then:

${\displaystyle \liminf _{n\to \infty }{\frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}\leq \liminf _{n\to \infty }{\frac {a_{n}}{b_{n}}}\leq \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}\leq \limsup _{n\to \infty }{\frac {a_{n+1}-a_{n}}{b_{n+1}-b_{n}}}.}$

Proof

Instead of proving the previous statement, we shall prove a slightly different one; first we introduce a notation: let ${\displaystyle (a_{n})_{n\geq 1}}$ be any sequence, its partial sum will be denoted by ${\displaystyle A_{n}:=\sum _{m\geq 1}^{n}a_{m}}$. The equivalent statement we shall prove is:

Let ${\displaystyle (a_{n})_{n\geq 1},(b_{n})_{\geq 1}}$ be any two sequences of real numbers such that

• ${\displaystyle b_{n}>0,\quad \forall n\in {\mathbb {Z} }_{>0}}$,
• ${\displaystyle \lim _{n\to \infty }B_{n}=+\infty }$,

then

${\displaystyle \liminf _{n\to \infty }{\frac {a_{n}}{b_{n}}}\leq \liminf _{n\to \infty }{\frac {A_{n}}{B_{n}}}\leq \limsup _{n\to \infty }{\frac {A_{n}}{B_{n}}}\leq \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}.}$

Proof of the equivalent statement

First we notice that:

• ${\displaystyle \liminf _{n\to \infty }{\frac {A_{n}}{B_{n}}}\leq \limsup _{n\to \infty }{\frac {A_{n}}{B_{n}}}}$ holds by definition of limit superior and limit inferior;
• ${\displaystyle \liminf _{n\to \infty }{\frac {a_{n}}{b_{n}}}\leq \liminf _{n\to \infty }{\frac {A_{n}}{B_{n}}}}$ holds if and only if ${\displaystyle \limsup _{n\to \infty }{\frac {A_{n}}{B_{n}}}\leq \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}}$ because ${\displaystyle \liminf _{n\to \infty }x_{n}=-\limsup _{n\to \infty }(-x_{n})}$ for any sequence ${\displaystyle (x_{n})_{n\geq 1}}$.

Therefore we need only to show that ${\displaystyle \limsup _{n\to \infty }{\frac {A_{n}}{B_{n}}}\leq \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}}$. If ${\displaystyle L:=\limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=+\infty }$ there is nothing to prove, hence we can assume ${\displaystyle L<+\infty }$ (it can be either finite or ${\displaystyle -\infty }$). By definition of ${\displaystyle \limsup }$, for all ${\displaystyle l>L}$ there is a natural number ${\displaystyle \nu >0}$ such that

${\displaystyle {\frac {a_{n}}{b_{n}}}<l,\quad \forall n>\nu .}$

We can use this inequality so as to write

${\displaystyle A_{n}=A_{\nu }+a_{\nu +1}+\dots +a_{n}<A_{\nu }+l(B_{n}-B_{\nu }),\quad \forall n>\nu ,}$

Because ${\displaystyle b_{n}>0}$, we also have ${\displaystyle B_{n}>0}$ and we can divide by ${\displaystyle B_{n}}$ to get

${\displaystyle {\frac {A_{n}}{B_{n}}}<{\frac {A_{\nu }-lB_{\nu }}{B_{n}}}+l,\quad \forall n>\nu .}$

Since ${\displaystyle B_{n}\to +\infty }$ as ${\displaystyle n\to +\infty }$, the sequence

${\displaystyle {\frac {A_{\nu }-lB_{\nu }}{B_{n}}}\to 0{\text{ as }}n\to +\infty {\text{ (keeping }}\nu {\text{ fixed)}},}$

and we obtain

${\displaystyle \limsup _{n\to \infty }{\frac {A_{n}}{B_{n}}}\leq l,\quad \forall l>L,}$

By definition of least upper bound, this precisely means that

${\displaystyle \limsup _{n\to \infty }{\frac {A_{n}}{B_{n}}}\leq L=\limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}},}$

and we are done.

Proof of the original statement

Now, take ${\displaystyle (a_{n}),(b_{n})}$ as in the statement of the general form of the Stolz-Cesàro theorem and define

${\displaystyle \alpha _{1}=a_{1},\alpha _{k}=a_{k}-a_{k-1},\,\forall k>1\quad \beta _{1}=b_{1},\beta _{k}=b_{k}-b_{k-1}\,\forall k>1}$

since ${\displaystyle (b_{n})}$ is strictly monotone (we can assume strictly increasing for example), ${\displaystyle \beta _{n}>0}$ for all ${\displaystyle n}$ and since ${\displaystyle b_{n}\to +\infty }$ also ${\displaystyle \mathrm {B} _{n}=b_{1}+(b_{2}-b_{1})+\dots +(b_{n}-b_{n-1})=b_{n}\to +\infty }$, thus we can apply the theorem we have just proved to ${\displaystyle (\alpha _{n}),(\beta _{n})}$ (and their partial sums ${\displaystyle (\mathrm {A} _{n}),(\mathrm {B} _{n})}$)

${\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\limsup _{n\to \infty }{\frac {\mathrm {A} _{n}}{\mathrm {B} _{n}}}\leq \limsup _{n\to \infty }{\frac {\alpha _{n}}{\beta _{n}}}=\limsup _{n\to \infty }{\frac {a_{n}-a_{n-1}}{b_{n}-b_{n-1}}},}$

which is exactly what we wanted to prove.

References

• Mureşan, Marian (2008), A Concrete Approach to Classical Analysis, Berlin: Springer, pp. 85–88, ISBN 978-0-387-78932-3.
• Stolz, Otto (1885), Vorlesungen über allgemeine Arithmetik: nach den Neueren Ansichten, Leipzig: Teubners, pp. 173–175.
• Cesàro, Ernesto (1888), "Sur la convergence des séries", Nouvelles annales de mathématiques, Series 3, 7: 49–59.
• Pólya, George; Szegő, Gábor (1925), Aufgaben und Lehrsätze aus der Analysis, I, Berlin: Springer.
• A. D. R. Choudary, Constantin Niculescu: Real Analysis on Intervals. Springer, 2014, ISBN 9788132221487, pp. 59-62
• J. Marshall Ash, Allan Berele, Stefan Catoiu: Plausible and Genuine Extensions of L’Hospital's Rule. Mathematics Magazine, Vol. 85, No. 1 (February 2012), pp. 52–60 (JSTOR)

External links

• l'Hôpital's rule and Stolz-Cesàro theorem at imomath.com
• Proof of Stolz–Cesàro theorem at PlanetMath.

Notes

1. https://www.springer.com/gp/book/9788132221470 see pages 59-60
2. l'Hôpital's rule and Stolz-Cesàro theorem at imomath.com

This article incorporates material from Stolz-Cesaro theorem on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.