Stolz-Cesàro Theorem

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

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

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:

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:

Hence by the definition of convergent sequences the result follows.

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.