Stolz–Cesàro theorem

Template:Short description In mathematics, the Stolz–Cesàro theorem is a criterion for proving the convergence of a sequence. It 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 Template:Math case

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

lim_{n \to \infty} \frac{a_{n + 1} - a_{n}}{b_{n + 1} - b_{n}} = l .

Then, the limit

lim_{n \to \infty} \frac{a_{n}}{b_{n}} = l .

Statement of the theorem for the Template:Math case

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

lim_{n \to \infty} \frac{a_{n + 1} - a_{n}}{b_{n + 1} - b_{n}} = l,

then

lim_{n \to \infty} \frac{a_{n}}{b_{n}} = l .

^[1]

Proofs

Proof of the theorem for the Template:Math case

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

| \frac{a_{n + 1} - a_{n}}{b_{n + 1} - b_{n}} - l | < \frac{ϵ}{2},

which is to say

l - ϵ / 2 < \frac{a_{n + 1} - a_{n}}{b_{n + 1} - b_{n}} < l + ϵ / 2, \forall n > ν .

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

(l - ϵ / 2) (b_{n + 1} - b_{n}) < a_{n + 1} - a_{n} < (l + ϵ / 2) (b_{n + 1} - b_{n}), \forall n > ν

.

Next we notice that

a_{n} = [(a_{n} - a_{n - 1}) + \dots + (a_{ν + 2} - a_{ν + 1})] + a_{ν + 1}

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

\begin{matrix} (l - ϵ / 2) (b_{n} - b_{ν + 1}) + a_{ν + 1} = (l - ϵ / 2) [(b_{n} - b_{n - 1}) + \dots + (b_{ν + 2} - b_{ν + 1})] + a_{ν + 1} < a_{n} \\ a_{n} < (l + ϵ / 2) [(b_{n} - b_{n - 1}) + \dots + (b_{ν + 2} - b_{ν + 1})] + a_{ν + 1} = (l + ϵ / 2) (b_{n} - b_{ν + 1}) + a_{ν + 1} . \end{matrix}

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

(l - ϵ / 2) + \frac{a_{ν + 1} - b_{ν + 1} (l - ϵ / 2)}{b_{n}} < \frac{a_{n}}{b_{n}} < (l + ϵ / 2) + \frac{a_{ν + 1} - b_{ν + 1} (l + ϵ / 2)}{b_{n}} .

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

c_{n}^{\pm} := \frac{a_{ν + 1} - b_{ν + 1} (l \pm ϵ / 2)}{b_{n}}

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

\begin{matrix} | c_{n}^{+} | < ϵ / 2, \forall n > n_{+}, \\ | c_{n}^{-} | < ϵ / 2, \forall n > n_{-}, \end{matrix}

therefore

l - ϵ < l - ϵ / 2 + c_{n}^{-} < \frac{a_{n}}{b_{n}} < l + ϵ / 2 + c_{n}^{+} < l + ϵ, \forall n > max {ν, n_{\pm}} = : N > 0,

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

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

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

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

a_{n} > 2 M (b_{n} - b_{ν + 1}) + a_{ν + 1}, \forall n > ν,

and

\frac{a_{n}}{b_{n}} > 2 M + \frac{a_{ν + 1} - 2 M b_{ν + 1}}{b_{n}}, \forall n > max {ν, n_{0}} .

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

c_{n} := \frac{a_{ν + 1} - 2 M b_{ν + 1}}{b_{n}}

is infinitesimal, thus

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

combining this inequality with the previous one we conclude

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

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

Proof of the theorem for the Template:Math case

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

a_{n} = (a_{n} - a_{n + 1}) + \dots + (a_{n + ν - 1} - a_{n + ν}) + a_{n + ν},

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

\begin{matrix} (l - ϵ / 2) (b_{n} - b_{n + ν}) + a_{n + ν} = (l - ϵ / 2) [(b_{n} - b_{n + 1}) + \dots + (b_{n + ν - 1} - b_{n + ν})] + a_{n + ν} < a_{n} \\ a_{n} < (l + ϵ / 2) [(b_{n} - b_{n + 1}) + \dots + (b_{n + ν - 1} - b_{n + ν})] + a_{n + ν} = (l + ϵ / 2) (b_{n} - b_{n + ν}) + a_{n + ν} . \end{matrix}

The two sequences

c_{ν}^{\pm} := \frac{a_{n + ν} - b_{n + ν} (l \pm ϵ / 2)}{b_{n}}

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

\begin{matrix} | c_{ν}^{+} | < ϵ / 2, \forall ν > ν_{+}, \\ | c_{ν}^{-} | < ϵ / 2, \forall ν > ν_{-}, \end{matrix}

thus, choosing $ν$ appropriately (which is to say, taking the limit with respect to $ν$ ) we obtain

l - ϵ < l - ϵ / 2 + c_{ν}^{-} < \frac{a_{n}}{b_{n}} < l + ϵ / 2 + c_{ν}^{+} < l + ϵ, \forall n > n_{0}

which concludes the proof.

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

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

Therefore, for each $ν > 0,$

\frac{a_{n}}{b_{n}} > 2 M + \frac{a_{n + ν} - 2 M b_{n + ν}}{b_{n}}, \forall n > n_{0} .

The sequence

c_{ν} := \frac{a_{n + ν} - 2 M b_{n + ν}}{b_{n}}

converges to $0$ (keeping $n$ fixed). Hence

\forall M > 0 \exists \bar{ν} > 0

such that

- M < c_{ν} < M, \forall ν > \bar{ν},

and, choosing $ν$ conveniently, we conclude the proof

\frac{a_{n}}{b_{n}} > 2 M + c_{ν} > M, \forall n > n_{0} .

Applications and examples

The theorem concerning the Template:Math case has a few notable consequences which are useful in the computation of limits.

Arithmetic mean

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

a_{n} := \sum_{m = 1}^{n} x_{m} = x_{1} + \dots + x_{n}, b_{n} := n

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

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

lim_{n \to \infty} \frac{x_{1} + \dots + x_{n}}{n} = lim_{n \to \infty} x_{n} .

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

exists (finite or infinite), then

$lim_{n \to \infty} \frac{x_{1} + \dots + x_{n}}{n} = lim_{n \to \infty} x_{n} .$

Geometric mean

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

a_{n} := \log (x_{1} \dots x_{n}), b_{n} := n,

again we compute

lim_{n \to \infty} \frac{a_{n + 1} - a_{n}}{b_{n + 1} - b_{n}} = lim_{n \to \infty} \log (\frac{x_{1} \dots x_{n + 1}}{x_{1} \dots x_{n}}) = 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

lim_{n \to \infty} \frac{\log (x_{1} \dots x_{n})}{n} = lim_{n \to \infty} \log ((x_{1} \dots x_{n})^{\frac{1}{n}}) = \log (l),

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

lim_{n \to \infty} \sqrt[n]{x_{1} \dots x_{n}} = lim_{n \to \infty} x_{n}

.

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

exists (finite or infinite), then

$lim_{n \to \infty} \sqrt[n]{x_{1} \dots x_{n}} = lim_{n \to \infty} x_{n} .$

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

lim_{n \to \infty} \sqrt[n]{y_{n}},

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

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

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 $(y_{n})_{n \geq 1}$ of (strictly) positive real numbers, suppose that
$lim_{n \to \infty} \frac{y_{n + 1}}{y_{n}}$

exists (finite or infinite), then

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

Examples

Example 1

lim_{n \to \infty} \sqrt[n]{n} = lim_{n \to \infty} \frac{n + 1}{n} = 1.

Example 2

\begin{matrix} 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{matrix}

where we used the representation of $e$ as the limit of a sequence.

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 $(a_{n})_{n \geq 1}$ and $(b_{n})_{n \geq 1}$ are two sequences such that $(b_{n})_{n \geq 1}$ is monotone and unbounded, then:

\underset{n \to \infty}{lim inf} \frac{a_{n + 1} - a_{n}}{b_{n + 1} - b_{n}} \leq \underset{n \to \infty}{lim inf} \frac{a_{n}}{b_{n}} \leq \underset{n \to \infty}{lim sup} \frac{a_{n}}{b_{n}} \leq \underset{n \to \infty}{lim sup} \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 $(a_{n})_{n \geq 1}$ be any sequence, its partial sum will be denoted by $A_{n} := \sum_{m \geq 1}^{n} a_{m}$ . The equivalent statement we shall prove is:

Let $(a_{n})_{n \geq 1}, (b_{n})_{\geq 1}$ be any two sequences of real numbers such that
$b_{n} > 0, \forall n \in ℤ_{> 0}$ ,

$lim_{n \to \infty} B_{n} = + \infty$ ,

then

$\underset{n \to \infty}{lim inf} \frac{a_{n}}{b_{n}} \leq \underset{n \to \infty}{lim inf} \frac{A_{n}}{B_{n}} \leq \underset{n \to \infty}{lim sup} \frac{A_{n}}{B_{n}} \leq \underset{n \to \infty}{lim sup} \frac{a_{n}}{b_{n}} .$

Proof of the equivalent statement

First we notice that:

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

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

\frac{a_{n}}{b_{n}} < l, \forall n > ν .

We can use this inequality so as to write

A_{n} = A_{ν} + a_{ν + 1} + \dots + a_{n} < A_{ν} + l (B_{n} - B_{ν}), \forall n > ν,

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

\frac{A_{n}}{B_{n}} < \frac{A_{ν} - l B_{ν}}{B_{n}} + l, \forall n > ν .

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

\frac{A_{ν} - l B_{ν}}{B_{n}} \to 0 as n \to + \infty (keeping ν fixed),

and we obtain

\underset{n \to \infty}{lim sup} \frac{A_{n}}{B_{n}} \leq l, \forall l > L,

By definition of least upper bound, this precisely means that

\underset{n \to \infty}{lim sup} \frac{A_{n}}{B_{n}} \leq L = \underset{n \to \infty}{lim sup} \frac{a_{n}}{b_{n}},

and we are done.

Proof of the original statement

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

α_{1} = a_{1}, α_{k} = a_{k} - a_{k - 1}, \forall k > 1 β_{1} = b_{1}, β_{k} = b_{k} - b_{k - 1} \forall k > 1

since $(b_{n})$ is strictly monotone (we can assume strictly increasing for example), $β_{n} > 0$ for all $n$ and since $b_{n} \to + \infty$ also $B_{n} = b_{1} + (b_{2} - b_{1}) + \dots + (b_{n} - b_{n - 1}) = b_{n} \to + \infty$ , thus we can apply the theorem we have just proved to $(α_{n}), (β_{n})$ (and their partial sums $(A_{n}), (B_{n})$ )

\underset{n \to \infty}{lim sup} \frac{a_{n}}{b_{n}} = \underset{n \to \infty}{lim sup} \frac{A_{n}}{B_{n}} \leq \underset{n \to \infty}{lim sup} \frac{α_{n}}{β_{n}} = \underset{n \to \infty}{lim sup} \frac{a_{n} - a_{n - 1}}{b_{n} - b_{n - 1}},

which is exactly what we wanted to prove.

References

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A. D. R. Choudary, Constantin Niculescu: Real Analysis on Intervals. Springer, 2014, Template:Isbn, 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

Notes

Template:PlanetMath attribution

[1] Template:Cite book

[2] 'Hôpital's rule and Stolz-Cesàro theorem at imomath.com

[1]

[2]

Stolz–Cesàro theorem

Contents

Statement of the theorem for the Template:Math case

Statement of the theorem for the Template:Math case

Proofs

Proof of the theorem for the Template:Math case

Proof of the theorem for the Template:Math case

Applications and examples

Arithmetic mean

Geometric mean

Examples

Example 1

Example 2

History

The general form

Statement

Proof

Proof of the equivalent statement

Proof of the original statement

References

External links

Notes

Navigation menu

Stolz–Cesàro theorem

Statement of the theorem for the Template:Math case

Statement of the theorem for the Template:Math case

Proofs

Proof of the theorem for the Template:Math case

Proof of the theorem for the Template:Math case

Applications and examples

Arithmetic mean

Geometric mean

Examples

Example 1

Example 2

History

The general form

Statement

Proof

Proof of the equivalent statement

Proof of the original statement

References

External links

Notes

Navigation menu

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