The Stolz–Cesàro theorem

In connectoin with the operation with sequences having limits (finite or infinite) we ran into the problem of "eliminating the indeterminates". We known that if a_n ⟶ ℓ and b_n ⟶ ℓ' (with ℓ' ≠ 0) we have

It may happen that a_n ⟶ 0 and b_n ⟶ 0, but the sequence {a_n/b_n} is still convergent. For example this is the case for a_n = b_n = 1/n

The following Stolz–Cesàro theorem has been called the de L'Hôpital's Theorem for sequences because is its analogous for the discrete case. It is due to Otto Stolz (1859-1906) and Ernesto Cesàro (1859-1906).

Theorem 2.7.1. (Stolz–Cesàro theorem) Let {a_n} and {b_n} be two sequences of real numbers convergent to 0, with {b_n} strictly positive, increasing and unbounded, if

$$\underset{n\to \mathrm{\infty }}{lim}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\ell$$

Then

$$\underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}=\ell$$

Proof. Fix ε > 0. There exists n₀ sucht that for n ≥ n₀

$$\ell -\epsilon <\frac{a_{n+1}-a_n}{b_{n+1}-b_n}<\ell +\epsilon$$

Because b_n+1 − b_n ≥ 0, this is equivalent to

Let k ≥ n₀. Summing the inequality from n₀, n₀ + 1, ..., k we get

$$\ell -\epsilon \sum _{n=n_0}^k{b}_{n+1}-b_n<\sum _{n=n_0}^k{a}_{n+1}-a_n<\ell -\epsilon \sum _{n=n_0}^k{b}_{n+1}-b_n$$

which gives

Dividing by b_{k + 1}, we obtain

Since lim_{k ⟶ ∞} b_k = ∞, the above inequality immediately gives

hence, lim_{n ⟶ ∞} a_n/b_n = ℓ.

If ℓ = +∞, then for all ε > 0 there exists an n₀ such that for all n ≥ n₀ we have

Thus it follows that

Let k ≥ n₀ be a natural number. Adding the above inequality for n₀, n₀ + 1, ..., k we get

Dividing by b_{k + 1}, we obtain

Now since b_n0/b_{k + 1} and a_n0/b_{k + 1} tend to zero as k ⟶ ∞, we find a_k+1/b_{k + 1} > ε, which implies that lim_{n ⟶ ∞} a_n/b_n = ∞.  □

If {b_n} is decreasing it suffices to change sign to the two sequences as {−b_n} and {−a_n} .

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