RAABE DUHAMEL’S CONVERGENCE TEST February 3, 2025
RAABE DUHAMEL TEST FOR
CONVERGENCE
1 Preliminary Definitions
Definition 1.1 Sequence: A sequence is a collection of numbers arranged
in a specific order, often following a rule.
It is used for defining patterns.
Definition 1.2 Series: A series is the sum of the terms of a sequence.
2 Ratio Test For Convergence
Definition 2.1 The Ratio Test is a test for convergence/divergence of
an infinite series.
Given an infinite series
X∞
an
n=1
, an represents the terms of a series.
Let
an+1
L = lim
n−>∞ an
L < 1 => absolute convergence
L > 1 => divergence
L = 1 => inconclusive results, hence requires other methods of proof.
2.1 Why It Works
The ratio test compares the growth rate of consecutive terms in the series.
If the terms shrink rapidly, i.eL < 1, then the series converges.
If the terms do not grow/shrink fast enough ie.L > 1, then the series diverges.
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RAABE DUHAMEL’S CONVERGENCE TEST February 3, 2025
3 Raabe Duhamel’s Test
Definition 3.1 Raabe’s test is an extension of the ratio test which can
be used in the event of an inconclusive result.
It tests convergence of an infinite series when the ration test fails.
Given a sequence ak assume that
ak+1
lim =1
k−>∞ ak
.
We solve for L, where
ak+1
L = lim k 1 −
k−>∞ ak
From the value of L, there are 3 conclusions that can be made:
L > 1 => convergence
L < 1 => divergence
L = 1 => inconclusion
3.1 Proof
Suppose that
ak+1
lim =1
k−>∞ ak
and
ak+1
L = lim k 1 − >1
k−>∞ ak
hence we can find R ∈ (1, L)
Take N ∈ N such that if
ak+1
n ≥ N, R < k 1 −
ak
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RAABE DUHAMEL’S CONVERGENCE TEST February 3, 2025
We know that lim gets closer and closer to L, so its bounded below by R.
k−>∞
L−R
Take ϵ = 2
ak+1 R ak+1
R<k 1− => <1−
ak k ak
−R
ak+1 R 1
=> <1− ≤ 1+
ak k k
−R
1 k+1
But 1 + =
k k
(k + 1)−R
=
k
(k + 1)−R
is the ratio of the terms from ∞ −R
P
k=1 n which converges by the
k
p-series test.
Examples
1. Given ∞
X 1.4.7...(3n + 1)
n=1
n2 .3n .n!
, show whether the series converges.
Here, there’s a p-series, geometric series and a facto-
rial, a good sign to use the Raabe’s test.
soln
1.4.7...(3n + 1)
Let an =
n2 .3n .n!
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RAABE DUHAMEL’S CONVERGENCE TEST February 3, 2025
an+1
STEP 1: find
an
an+1 1.4.7...(3n + 1)(3n + 4) 1.4.7...(3n + 1)
=
an (n + 1)2 .3(n+1) .(n + 1)! n2 .3n .n!
n2 (3n + 4)
=
3.(n + 1)3
3n3 + 4n2
= 3
3n + 9n2 + 9n + 3
an+1
STEP 2: take lim to justify the use of Raabe’s
n−>∞ an
test
an+1 3n3 + 4n2
lim = lim =1
n−>∞ an n−>∞ 3n3 + 9n2 + 9n + 3
an+1
STEP 3: calculate lim n 1 − an
, and deduce your
n−>∞
conclusion
3n3 + 4n2
an+1
lim n 1 − = lim n 1 − 3
n−>∞ an n−>∞ 3n + 9n2 + 9n + 3
3
3n + 9n2 + 9n + 3 3n3 + 4n2
= lim n −
n−>∞ 3n3 + 9n2 + 9n + 3 3n3 + 9n2 + 9n + 3
5n3 + 9n2 + 3n
= lim
n−>∞ 3n3 + 9n2 + 9n + 3
5
= >1
3
Hence the series converges.
2. Show whether the given series converges;
∞
X 1 3 5 2k − 1
, , , ...,
k=1
4 6 8 2k + 2
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RAABE DUHAMEL’S CONVERGENCE TEST February 3, 2025
soln
∞
X 1 3 5 2k − 1
Let ak = , , , ...,
k=1
4 6 8 2k + 2
∴ We have that
ak+1 1 3 5 2k − 1 2k 1 3 5 2k − 1
= , , , ..., , ÷ , , , ...,
ak 4 6 8 2k + 2 2k + 3 4 6 8 2k + 2
ak+1 2k
=
ak 2k + 3
ak+1 2k
lim = lim
k−>∞ ak k−>∞ 2k + 3
2
= lim
k−>∞ 2 + 3
k
=1
ak+1 2k
k 1− =k 1−
ak 2k + 3
2k + 3 − 2k
=k
2k + 3
3
=k
2k + 3
3k
=
2k + 3
ak+1 3k
lim k 1 − = lim
k−>∞ ak k−>∞ 2k + 3
3
= lim
k−>∞ 2 + k3
3
= >1
2
Hence the series converges.
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RAABE DUHAMEL’S CONVERGENCE TEST February 3, 2025
References
1. Find the YouTube video used
2. How do I state my reference is from ChatGPT
3. Also state the other YouTube video you didn’t use, just for the numbers
Likely Questions
1. O00 ‘ne question to look up, what do you do when the Raabe’s test is
inconclusive?
2. Can you use this test for a finite series?
3. Do finite series always converge?
ASK DR KEN FOR THE TEMPLATE
TAKE NOTE OF THE REMARKS DR KEN MENTIONED.
