Download Absolute Convergence: Tests for Absolute and Conditional Convergence and more Study notes Advanced Calculus in PDF only on Docsity!
Absolute convergence
Definition A series
P
an is called absolutely convergent if the series of absolute values
P
|an| is convergent. If the terms of the series an are positive, absolute convergence is the same as convergence.
Example Are the following series absolutely convergent? X^ ∞
n=
(−1)n+ n^3
X^ ∞
n=
(−1)n n
I (^) To check if the series P∞ n=1(−1) n 3 n +1is absolutely convergent, we need to check if the series of absolute values
P∞
n=
1 n^3 is convergent. I (^) Since
P∞
n=
1 P^ n^3 is a p-series with^ p^ = 3^ >^ 1, it converges and therefore ∞ n=
(−1)n+ n^3 is absolutely convergent. I (^) To check if the series
P∞
n=
(−1)n n is absolutely convergent, we need to check if the series of absolute values
P∞
n=
1 n is convergent. I (^) Since P∞ n=1^1 n is a p-series with p = 1, it diverges and therefore P∞ n=
(−1)n n is not absolutely convergent.
Conditional convergence
Definition A series
P
an is called conditionally convergent if the series is convergent but not absolutely convergent. Which of the series in the above example is conditionally convergent? I (^) Since the series P∞ n=1(−1)
n+ n^3 is absolutely convergent, it is^ not conditionally convergent. I (^) Since the series
P∞
n=
(−1)n n is convergent (used the alternating series test last day to show this), but the series of absolute values
P∞
n=
1 n is not convergent, the series
P∞
n=
(−1)n n is conditionally convergent.
The Ratio Test
This test is useful for determining absolute convergence. Let
P∞
n=1 an^ be a series (the terms may be positive or negative). Let L = limn→∞
˛ an a+1n
I (^) If L < 1, then the series P∞ n=1 an converges absolutely (and hence is convergent). I (^) If L > 1 or ∞, then the series P∞ n=1 an is divergent. I (^) If L = 1, then the Ratio test is inconclusive and we cannot determine if the series converges or diverges using this test. This test is especially useful where factorials and powers of a constant appear in terms of a series. (Note that when the ratio test is inconclusive for an alternating series, the alternating series test may work. ) Example 1 Test the following series for convergence X^ ∞
n=
(−1)n−^1
2 n n!
I (^) limn→∞
˛ an a+1n
˛ = limn→∞
2 n+
(n+1)! 2 n
n!
˛ = limn→∞ (^) n+1^2 = 0 < 1.
I (^) Therefore, the series converges.
Example 2
Ratio Test Let
P∞
n=1 an^ be a series (the terms may be positive or negative). Let L = limn→∞
˛ an a+1n
If L < 1, then the series
P∞
n=1 an^ converges absolutely. If L > 1 or ∞, then the series
P∞
n=1 an^ is divergent. If L = 1, then the Ratio test is inconclusive. Example 2 Test the following series for convergence
X^ ∞
n=
(−1)n
“ (^) n 5 n
I (^) limn→∞
˛˛ an+ an
˛˛ = limn→∞
˛˛ (n+1)
5 n+ n
5 n
˛˛ = limn→∞ n+ 5 n = 1 5 limn→∞(1 + 1/n) =^
1 5 <^ 1. I (^) Therefore, the series converges.
Example 4
Example 4 Test the following series for convergence
P∞
n=
(−1)n n^2 I (^) We know already that this series converges absolutely and therefore it converges. (we could also use the alternating series test to deduce this). I (^) Lets see what happens when we apply the ratio test here.
I (^) limn→∞
˛ an a+1n
˛ = limn→∞
1
(n+1)^2 1
n^2
˛ = limn→∞
n n+
limn→∞
1 1+1/n
I (^) Therefore the ratio test is inconclusive here.
The Root Test
Root Test Let
P∞
n=1 an^ be a series (the terms may be positive or negative). I (^) If limn→∞ n
p |an| = L < 1, then the series
P∞
n=1 an^ converges absolutely (and hence is convergent). I (^) If limn→∞ n
p |an| = L > 1 or limn→∞ n
p |an| = ∞, then the series
P∞
n=1 an is divergent. I (^) If limn→∞ n
p |an| = 1, then the Root test is inconclusive and we cannot determine if the series converges or diverges using this test.
Example 5 Test the following series for convergence
P∞
n=1(−1)
n− 1
2 n n+
”n
I (^) limn→∞ n
p |an| = limn→∞ n
r“ 2 n n+
”n = limn→∞ (^) n^2 +1n = limn→∞ (^) 1+1^2 /n = 2 > 1 I (^) Therefore by the n th root test, the series P∞ n=1(−1)n−^1
2 n n+
”n diverges.
Example 7
Root Test For
P∞
n=1 an.^ L^ = limn→∞^
pn|an|.
If L < 1, then the series
P∞
n=1 an^ converges absolutely. If L > 1 or ∞, then the series
P∞
n=1 an^ is divergent. If L = 1, then the Root test is inconclusive. Example 7 Test the following series for convergence
P∞
n=
ln n n
”n .
I (^) limn→∞ n
p |an| = limn→∞ n
r“ ln n n
”n = limn→∞ lnn^ n = limx→∞ lnx^ x =
(L′Hop) limx→∞ 1 / 1 x = 0 < 1
I (^) Therefore by the n th root test, the series P∞ n=
ln n n
”n converges.
Rearranging sums
If we rearrange the terms in a finite sum, the sum remains the same. This is not always the case for infinite sums (infinite series). It can be shown that: I (^) If a series P^ an is an absolutely convergent series with P^ an = s, then any rearrangement of
P
an is convergent with sum s. I (^) It a series P^ an is a conditionally convergent series, then for any real number r , there is a rearrangement of
P
an which has sum r.
I (^) Example The series P∞ n=1(− 2 1)n nis absolutely convergent with P∞ n=
(−1)n 2 n^ =^
2 3 and hence any rearrangement of the terms has sum^
2
