Stolz-Cesàro Theorem

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\left({a_n}\right)$ be any sequence and $\left({b_n}\right)$ a sequence of positive real numbers such that $\displaystyle \sum_{i \mathop = 0}^{\infty}b_n = \infty$.

If:

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

then also:

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


Corollary

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

If:

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

then also:

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


Proof

Define the following sums:

$\displaystyle A_n = \sum_{i \mathop = 1}^{n}a_i$
$\displaystyle B_n = \sum_{i \mathop = 1}^{n}b_i$

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

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

$\displaystyle \left({L - \mu}\right) b_n < a_n < \left({L + \mu}\right) b_n$ for all $n > k$

Rewrite the sum as:

\(\displaystyle A_n = a_1 + a_2 + \cdots + a_k + a_{k+1} + \cdots + a_n\) \(\) \(\displaystyle \)
\(\displaystyle \implies \ \ \) \(\displaystyle a_1 + a_2 + \cdots + a_k + \left({L - \mu}\right)\left({b_{k+1} + \cdots + b_n}\right)\) \(<\) \(\, \displaystyle a_1 + a_2 + \cdots + a_n \, \) \(\, \displaystyle <\, \) \(\displaystyle a_1 + a_2 + \cdots + a_k + \left({L + \mu}\right)\left({b_{k+1} + \cdots + b_n}\right)\)
\(\displaystyle \implies \ \ \) \(\displaystyle A_k + \left({L - \mu}\right)\left({B_n - B_k}\right)\) \(<\) \(\, \displaystyle A_n \, \) \(\, \displaystyle <\, \) \(\displaystyle A_k + \left({L + \mu}\right)\left({B_n - B_k}\right)\)

Divide above by $B_n$:

$\displaystyle \frac{A_k + \left({L-\mu}\right) B_k}{B_n} + \left({L - \mu}\right) < \frac{A_n}{B_n} < \left({L + \mu}\right) + \frac{A_k + \left({L+\mu}\right) B_k}{B_n}$

Let $k$ be fixed.

From Reciprocal of Null Sequence and Combination Theorem for Sequences, sequence $\displaystyle \left({\frac{A_k + \left({L \pm \varepsilon}\right) B_k}{B_n}}\right)$ converges to zero.

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

$\displaystyle \left\vert{\frac{A_k + \left({L \pm \mu}\right) B_k}{B_n}}\right\vert < \mu$ for all $n > N$

Substitute the above into the inequation and obtain:

\(\displaystyle \) \(\) \(\displaystyle L - 2 \mu< \frac{A_n}{B_n} < L + 2 \mu\)
\(\displaystyle \iff\) \(\) \(\displaystyle \left\vert{\frac{A_n}{B_n} - L}\right\vert < \varepsilon\) for all $n > N$

Hence by the definition of convergent sequences the result follows.

$\blacksquare$


Remarks


  • 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.