WEEK2 CHAPTER 1 MAT455

1.3 Tests for Convergence

We have discussed the method of determining the convergence and divergence of a

particular series by first writing the nth partial sum, Sn, of the series in closed form,
and then investigate the lim𝑛→∞ 𝑆𝑛 .

However, some series cannot be written in the closed form. Thus, the following tests
are used to determine the convergence and divergence of these series.

A) The Divergence Test

Theorem 1
i) If lim𝑛→∞ 𝑎𝑛 ≠ 0 , the series ∑𝑛=∞
𝑛=1 𝑎𝑛 diverges.

ii) If lim𝑛→∞ 𝑎𝑛 = 0 , the series ∑𝑛=∞

𝑛=1 𝑎𝑛 may either converge or
diverge.

Note: if you can see at a glance that 𝑙𝑖𝑚𝑛→∞ 𝑎𝑛 ≠ 0, then the Divergence Test should be used

Example 1
Use the Divergence Test to determine if the following series is divergent.
1 𝑒𝑛
a) ∑∞
𝑛=1 2
𝑛
b) ∑∞
𝑛=1 𝑛 c) ∑∞
𝑛=1
2 𝑛

𝑛 𝑛2
d) ∑∞
𝑛=1 ln(𝑛+1) e) ∑∞
𝑛=1 5𝑛2 +4

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B) The Integral Test

Theorem 2
Suppose f is a continuous, positive, decreasing function on [1 , ∞) and let an =
f(n). Then the series ∑𝑛=∞
𝑛=1 𝑎𝑛 is convergent if and only if the improper
∞
integral ∫1 𝑓(𝑥)𝑑𝑥 is convergent. In other words

∞
i) If ∫1 𝑓(𝑥)𝑑𝑥 is convergent, then ∑𝑛=∞
𝑛=1 𝑎𝑛 is convergent.

∞
ii) If ∫1 𝑓(𝑥)𝑑𝑥 is divergent, then ∑𝑛=∞
𝑛=1 𝑎𝑛 is divergent.

∞
Note: i) an = f(n), where If ∫1 𝑓(𝑥)𝑑𝑥 is easily evaluated, then the Integral Test is effective.
∞ 𝑡
ii) If ∫𝑎 𝑓(𝑥)𝑑𝑥 = 𝑙𝑖𝑚𝑡→∞ ∫𝑎 𝑓(𝑥) 𝑑𝑥

Example 2
Use the Integral Test to determine if the following series is convergent or divergent.
1 ln 𝑛 𝑛 1
a) ∑∞ 𝑛=1 3 b) ∑∞𝑛=1 c) ∑∞
𝑛=1 𝑛 d) ∑∞
𝑛=1 (𝑛+1)
𝑛 𝑛 𝑒 ln(𝑛+1)

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C) The Comparison Test

- the idea is to compare a given series with a series that is known be
convergent or divergent (eg: p-series, geometric series)

Theorem 3
Suppose that ∑ 𝑎𝑛 and ∑ 𝑏𝑛 are series with positive terms.

i) If ∑ 𝑏𝑛 is convergent and 𝑎𝑛 ≤ 𝑏𝑛 for all n, then ∑ 𝑎𝑛 is also convergent.

ii) If ∑ 𝑏𝑛 is divergent and 𝑎𝑛 ≥ 𝑏𝑛 for all n, then ∑ 𝑎𝑛 is also divergent.

Tips for applying the comparison test

1. Delete the constant in the denominator of an (as this will not affect the
convergence or divergence of the series).

2. If the numerator or the denominator of an appears in a polynomial form,

eliminate all except for the term with the highest power.

3. Make a guess as to whether the series converge or diverge.

- if the guess is convergence, then, choose a bigger series for comparison.

- if the guess is divergence, then choose a smaller series for comparison.

The Comparison Test works well for comparing “messy” algebraic series to a p-
series, geometric series or any other series that is known to converge. In choosing
an appropriate p-series, we must choose one with an nth term of the same
magnitude as the nth term of the given series.
1 1
Eg: 1. If ∑ 3𝑛2 −4𝑛+5 , compare with ∑ 𝑛2 .

1 1
2. If ∑ , compare with ∑ 𝑛1/2 .
√3𝑛−2

𝑛2 −10 1
3. If ∑ 4𝑛5 +𝑛3 , compare with ∑ 𝑛3 .

√𝑛 1
4. If ∑ √𝑛3 , compare with ∑ 𝑛 .
+3

1 1
5. If ∑ 2𝑛+1 , compare with ∑ 2𝑛 .

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Example 3
Use the Comparison Test to determine if the series converges or diverges.

1 5
a) ∑∞
𝑛=1 2+3𝑛 b) ∑∞
𝑛=1 2𝑛2 +4𝑛+3

5𝑛 +1 1
c) ∑∞
𝑛=1 2𝑛 −1 d) ∑∞
𝑛=1 2+ √𝑛

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D) The Limit Comparison Test

Theorem 4
Suppose that ∑ 𝑎𝑛 and ∑ 𝑏𝑛 are series with positive terms. If
𝑎
lim𝑛→∞ 𝑏𝑛 = 𝑐
𝑛

Where c is a finite number and c > 0, then either both series converge or both
diverge.

Example 4
Use the Limit Comparison Test to determine whether the series converge or diverge.

1 2𝑛2 + 3𝑛 1 1+ 2𝑛
a) ∑∞
𝑛=1 2𝑛 −1 b) ∑∞
𝑛=1 c) ∑∞
𝑛=1 √𝑛2 d) ∑∞
𝑛=1 1 + 3𝑛
√5 + 𝑛 5 +1

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E) The Ratio Test

Theorem 5
𝑎𝑛+1
Let ∑∞
𝑛=1 𝑎𝑛 be a series with positive terms such that lim𝑛→∞ | 𝑎 | = L
𝑛

i) If L < 1, then the series ∑∞

𝑛=1 𝑎𝑛 is convergent.

ii) If L > 1, then the series ∑∞

𝑛=1 𝑎𝑛 is divergent.

iii) If L = 1, then the Ratio test is inconclusive; that is no conclusion can be

drawn about the convergence or divergence of the series ∑∞ 𝑛=1 𝑎𝑛 .

Note: this test is usually used when 𝑎𝑛 involves factorials or n th power.

Example 5
Use the Ratio Test to determine if the following series is convergent or divergent.
𝑛2 1 (𝑛+1)! 𝑛! 2𝑛−1
a) ∑∞
𝑛=1 2𝑛 b) ∑∞
𝑛=1 (2𝑛)! c) ∑∞
𝑛=1 d) ∑∞
𝑛=1 (2𝑛+1)! e) ∑∞
𝑛=1 3𝑛 (𝑛+1)
2𝑛 𝑛2

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TRY THIS

Use any relevant test to determine whether the following series converges or diverges.
5𝑘 +2𝑘
1. ∑∞
𝑘=1 (Ans: Diverge)
3𝑘 +4

𝜋𝑘
2. ∑∞
𝑘=1 (Ans: Converge)
52𝑘 (3𝑘)!

2𝑘
3. ∑∞
𝑘=1 ln(𝑘+3) (Ans: Diverge)

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