Stolz–Cesàro theorem

Case 1: suppose ${\displaystyle (b_{n})}$ strictly increasing and divergent to ${\displaystyle +\infty }$, and ${\displaystyle -\infty <l<\infty }$. 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 ${\displaystyle b_{n}>0}$ 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 exists ${\displaystyle n_{\pm }>n_{0}>0}$, 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 ${\displaystyle 2M>0}$ there exists ${\displaystyle \nu >0}$ such that for all ${\displaystyle n>\nu }$

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

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

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

and

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

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

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

is infinitesimal, thus

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

combining this inequality with the previous one we conclude

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

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.

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

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

and for any ${\displaystyle \epsilon /2>0,}$ ${\displaystyle \exists n_{0}}$ such that for all ${\displaystyle n>n_{0}}$ we have

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

The two sequences

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

are infinitesimal since by hypothesis ${\displaystyle a_{n+\nu },b_{n+\nu }\to 0}$ as ${\displaystyle \nu \to \infty }$, thus for all ${\displaystyle \epsilon /2>0}$ there are ${\displaystyle \nu _{\pm }>0}$ such that

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

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

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

which concludes the proof.

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

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

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

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

The sequence

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

converges to ${\displaystyle 0}$ (keeping ${\displaystyle n}$ fixed). Hence

${\displaystyle \forall M>0\,~\exists {\bar {\nu }}>0}$ such that ${\displaystyle -M<c_{\nu }<M,\,\forall \nu >{\bar {\nu }},}$

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

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

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

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

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

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

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

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

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

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:

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

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

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

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.

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.