Elementary Convergence Tests

By comparing the terms of a given series with the terms of a series for which convergence or divergence is known, we are then able to determine whether the series converges too.

Theorem 3.2.1 (The Comparison Test) Let ∑ a_n and ∑ b_n two series of positive terms such that 0 ≤ a_n ≤ b_n, ∀n

∑ b_n convergent ⇒ ∑ a_n convergent.
∑ a_n divergent ⇒ ∑ b_n divergent.

The series ∑ b_n is said the majorant, of ∑ a_n which is said minorant.

Proof. Because the first ∑ b_n converges, it satisfies the Cauchy criterion for series. Hence given, given ε > 0, there is an N so large that if n ≥ m > N then

| \sum_{j = m + 1}^{m} a_{j} | < | \sum_{j = m + 1}^{m} b_{j} | < ε

thus Cauchy criterion is satisfied by ∑ a_n too. Therefore it converges □

Thereom 3.2.2 (The root test). Let ∑ a_n a positive term series. If

lim_{n \to \infty} \sqrt[n]{a_{n}} = ℓ < 1

then the series converges.

Proof. Consider first the case ℓ < 1. Since |(a_n)^1/n − ℓ| ≤ ε/2 for a given ε > 0, we have (a_n)^1/n ≤ ℓ + ε/2.
Since ℓ < 1, we have also that ℓ < 1 − ε for a suitable ε > 0. For this ε we have that

(a_n)^1/n ≤ ℓ + ε/2 < (1 − ε) + ε/2 = 1 − ε/2

Thus

a_n < (1 − ε/2)ⁿ

which by comparison with the geometric series Σ (1 − ε/2)ⁿ, we deduce that the original series converges.

If instead ℓ > 1, with a similar argument we deduce that

a_n > (1 − ε/2)ⁿ

for a give ε > 0. Thus a_n → ∞, and the series diverges. □

Thereom 3.2.3. (D'Alembert's ratio test) Let ∑ a_n a series of positive terms. If the following limit exists

lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = ℓ < 1

Then if ℓ < 1, the series converges; If ℓ > 1, the series diverges.

Proof. Let ℓ finite. By similar reasoning used to prove the root test we have for the case ℓ < 1

\frac{a_{n + 1}}{a_{n}} < (1 - \frac{ε}{2})

for some ε > 0. This implies by reasoning iteratively that

a_{n + 1} < (1 - \frac{ε}{2}) a_{n} < (1 - \frac{ε}{2}) \cdot (1 - \frac{ε}{2}) a_{n - 1} < \dots < {(1 - \frac{ε}{2})}^{n} a_{1}

By comparison with the geometric series ∑ (1 − ε/2)ⁿ a₁, the initial series converges.

If instead ℓ >1, with a similar argument we deduce that

a_{n + 1} > {(1 + \frac{ε}{2})}^{n} a_{1}

for a given, ε > 0. Thus a_n → +∞, and the series diverges. □

Thereom 3.2.4 (Asymptotic comparison test). Let {a_n} and {b_n} two Asymptotic positive sequences

a_n ~ b_n

then the correspondent series ∑ a_n and ∑ b_n, have the same behaviour, either they are both convergent or divergent.

Proof.. To say that a_n ~ b_n means that a_n/b_n → 1 for n → ∞. This implies that for every ε > 0 sia definitivamente 1 − ε < a_n/b_n < 1 + ε; for example taking (ε=1/2), we have

1/2 < a_n / b_n < 3/2

or being by hyphothesis b_n > 0

1/2 b_n < a_n < 3/2 b_n

By the comparison test, the first inequality implies that if ∑ a_n conveges, then also b_n converges; the second inequality implies that if ∑ a_n diverges, then also ∑ b_n diverges. This is precisely what is meant by "having the same behaviour". □

Remark. Note that in general the fact that ∑ a_n conveges, then also b_n converges, does not mean that the two series have the same sum: a_n ~ b_n says that the terms of the two series are "similar" but the value of a sum depends upon all terms of the sum.

Example 3.2.5 The series Σ 1/n² converge since 1/n² ~ 1/n(n + 1) and Mengoli's series Σ 1/n(n + 1) converges. ■

Example 3.2.6 (Generalized harmonic series) The generalized harmonic series is

\sum_{n = 1}^{\infty} \frac{1}{n^{α}}

if α ≥ 2 then the series converges by the comparison test 1/n^α ≤ 1/n² and Σ 1/n² converges. However if α ≤ 1 the series diverges by comparison with the harmonic series. It can be proved (we shall use integral calculus for the proof) that for the intermediate values 1 < α < 2, there's still convergence. To recap we have that

\sum_{n = 1}^{\infty} \frac{1}{n^{α}}, convergent if α > 1; divergent if α \leq 1

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