Stolz-Cesàro Theorem

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Theorem

Let $\sequence {a_n}$ be a sequence.

This article, or a section of it, needs explaining. In particular: Domain of $\sequence {a_n}$ -- $\R$ presumably but could it be $\C$? You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

Let $\sequence {b_n}$ be a sequence of (strictly) positive real numbers such that:

$\ds \sum_{i \mathop = 0}^\infty b_n = \infty$

If:

$\ds \lim_{n \mathop \to \infty} \dfrac {a_n} {b_n} = L \in \R$

then also:

$\ds \lim_{n \mathop \to \infty} \dfrac {a_1 + a_2 + \cdots + a_n} {b_1 + b_2 + \cdots + b_n} = L$

Corollary

Let $\sequence {a_n}$ and $\sequence {b_n}$ be sequences in $\R$ such that $\sequence {b_n}$ is strictly increasing and $\ds \lim_{n \mathop \to \infty} b_n = \infty$.

If:

$\ds \lim_{n \mathop \to \infty} \frac {a_n - a_{n - 1} } {b_n - b_{n - 1} } = L \in \R$

then also:

$\ds \lim_{n \mathop \to \infty} \frac {a_n} {b_n} = L$

Proof

Define the following sums:

$\ds A_n = \sum_{i \mathop = 1}^n a_i$

$\ds B_n = \sum_{i \mathop = 1}^n b_i$

Let $\epsilon > 0$ and $\mu = \dfrac {\epsilon} 2$.

By the definition of convergent sequences, there exists $k \in \N$ such that:

$\forall n > k: \paren {L - \mu} b_n < a_n < \paren {L + \mu} b_n$

Rewrite the sum as:

\(\ds A_n = a_1 + a_2 + \cdots + a_k + a_{k + 1} + \cdots + a_n\)

\(\)

\(\ds \)

\(\ds \leadsto \ \ \)

\(\ds a_1 + a_2 + \cdots + a_k + \paren {L - \mu} \paren {b_{k + 1} + \cdots + b_n}\)

\(<\)

\(\, \ds a_1 + a_2 + \cdots + a_n \, \)

\(\, \ds < \, \)

\(\ds a_1 + a_2 + \cdots + a_k + \paren {L + \mu} \paren {b_{k + 1} + \cdots + b_n}\)

\(\ds \leadsto \ \ \)

\(\ds A_k + \paren {L - \mu} \paren {B_n - B_k}\)

\(<\)

\(\, \ds A_n \, \)

\(\, \ds < \, \)

\(\ds A_k + \paren {L + \mu} \paren {B_n - B_k}\)

Divide above by $B_n$:

$\dfrac {A_k + \paren {L - \mu} B_k} {B_n} + \paren {L - \mu} < \dfrac {A_n} {B_n} < \paren {L + \mu} + \dfrac {A_k + \paren {L + \mu} B_k} {B_n}$

Let $k$ be fixed.

From Reciprocal of Null Sequence and Combination Theorem for Sequences, the sequence $\sequence {\dfrac {A_k + \paren {L \pm \epsilon} B_k} {B_n} }$ converges to zero.

By the definition of convergent sequences, there exists $N > k > 0$ such that:

$\size {\dfrac {A_k + \paren {L \pm \mu} B_k} {B_n} } < \mu$ for all $n > N$

Substitute the above into the inequality and obtain:

\(\ds \)

\(\)

\(\ds L - 2 \mu < \frac {A_n} {B_n} < L + 2 \mu\)

\(\ds \)

\(\leadstoandfrom\)

\(\ds \size {\frac {A_n} {B_n} - L} < \epsilon\)

for all $n > N$

Hence by the definition of convergent sequences the result follows.

$\blacksquare$

Remarks

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• Using the similar proof technique with limits inferior and superior, a more general version of this theorem can be obtained. In that case the limit $L$ can be either a real number or $\pm \infty$.
• By setting $b_n = 1$ the theorem turns into Cesàro Mean for real-valued sequences.

Source of Name

This entry was named for Otto Stolz and Ernesto Cesàro.