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8.4 OTHER CONVERGENCE TESTS
Alternating Series Test
Alternating Series Estimation Theorem
Ratio Tests
Root Tests
The Alternating Series Test
If the alternating series
∞
∑ (−1)
n =1
(bn > 0) satisfies
convergent.
n −1
bn = b1 – b2 + b3 – b4 +
⋅⋅⋅
i) bn+1 ≤ bn for all n
bn = 0 the series is
ii) nlim
→∞
“This test says that if the terms of an alternating series
decrease to 0 in absolute value, then the series
converges.”
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Example: Does the alternating harmonic series
∞
(-1)
n=1 n
∑
or diverge?
Ans. Convergent because the series satisfies both
conditions of the alternating series test.
Alternating Series Estimation Theorem
An alternating series with sum S may be approximated
by a using partial sum, sn . The following is true for an
alternating series that that satisfies the hypotheses of
the alternating series test:
a) S lies between any two successive partial sum; that is either
sn ≤ S ≤ sn+1
or
sn +1 ≤ S ≤ sn
depending on which partial sum is larger.
b) If S is approximated by sn , then the absolute error S-sn satisfies
S-sn ≤ an +1
In addition, the sign of the error S – sn is the same as that of
the coefficient of an+1
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∞
∑ (−1)
n +1
n =1
Sn
an = a1 − a2 + a3 − a4 + ⋅ ⋅ ⋅.
S1=a1
S1
S3 = S2 + a3
S3
S5
S7
a1
S8
S
Sn 6
S4
a2
a3
a4
a5
S4 = S3 – a4
S2
S2=S1 ‐a1
0
1
1 2 3 4 5 6 7 8 1
n
1 1 1
1 1 1
R = S-sn ≤ an +1
Example: Show that the series is convergent. How many
terms of the series do we need to add in order to find the
sum of the indicated accuracy? i.e. S − sn ≤ 10 −6
( −1)n +1
∑
n
n =1
∞
Ans. i )
1
1
1
<
, ii ) lim
=0
n +1 n
n →∞ n
About 1,000,000 terms to get
desired accuracy
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Find the number of terms that need to be added to
give the indicated accuracy.
b)
(-1)n+1
; ( error < 0.00005 )
n6
n=1
∞
∑
Ans. 5 terms.
Absolute Convergence
∞
A series
∑b
n
= b1 + b2 + ⋅⋅⋅bn + ⋅⋅⋅ is said to converge absolutely
n=1
∞
if the series of absolute values
∑b
n
= b1 + b2 + ⋅⋅⋅ + bn + ⋅⋅⋅ converges
n=1
and is said to diverge absolutely if the series of absolute values diverges.
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Conditionally Convergent
A series
a is called conditionally convergent if
∑
n
it is convergent but not absolutely convergent.
If a series
∑a
n
is absolutely convergent,
then it is convergent.
Discuss the convergence (absolute, conditional,
divergence) of the following:
a) 1 -
1 1
1
1 1
Ans
− 2 + 3 + 4 − 5 − ⋅⋅⋅ Ans.
2 2 2
2 2
1 1 1 1
b) 1- + − + + ⋅⋅⋅
2 3 4 5
Converges absolutely
Ans. The series of absolute values
is the divergent harmonic series.
It diverges absolutely. It is
conditionally convergent.
∞
c)
cos n
2
n=1 n
∑
The series of absolute values
Converges by the comparison
test. The series converges
absolutely and therefore
converges.
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The Ratio Test
i) if lim
n →∞
an +1
= L < 1, then the series
an
∞
∑a
n
is absolutely
n=1
convergent (and therefore convergent).
ii) If lim
n →∞
series
an +1
a
= L > 1, or lim n +1 = ∞, then the
n →∞ a
an
n
∞
∑a
n
is divergent.
n=1
iii) If lim
n →∞
an +1
= 1, the Ratio Test is inconclusive;
an
and no conclusion can be made about the convergence
or divergence of the series.
The Root Test
i) If lim
n
n →∞
an = L < 1, then the series,
∞
∑a
n
is absolutely
n=1
convergent
g
((and therefore convergent).
g )
ii) If lim n an = L > 1 or If lim n an = ∞, then the
n →∞
series
n →∞
∞
∑a
n
is divergent.
n=1
iii) If lim n an = 1, the Root Test is inconclusive.
n →∞
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Examples (a, b) Determine whether the series is
absolutely convergent, conditionally convergent, or
divergent.
( −1)5n −1
a) ∑
2 n +2
n =1 ( n + 1) 4
∞
( −1)n
b) ∑
n
n = 2 (ln n )
Ans. Since the limit from the
Ratio Test = 5/4 >1, the series
diverges.
∞
Ans. Converges absolutely by
the Root Test
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