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Absolute Convergence: Tests for Absolute and Conditional Convergence, Study notes of Advanced Calculus

The concept of absolute convergence of series and provides tests to determine absolute and conditional convergence. The Ratio Test and Root Test are introduced to determine the convergence behavior of series. The document also covers the importance of absolute convergence and its relationship to convergence.

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Absolute convergence
Definition A series Panis called absolutely convergent if the series of
absolute values P|an|is convergent.
If the terms of the series anare positive, absolute convergence is the same as
convergence.
Example Are the following series absolutely convergent?
X
n=1
(1)n+1
n3,
X
n=1
(1)n
n.
ITo check if the series P
n=1
(1)n+1
n3is absolutely convergent, we need to
check if the series of absolute values P
n=1
1
n3is convergent.
ISince P
n=1
1
n3is a p-series with p= 3 >1, it converges and therefore
P
n=1
(1)n+1
n3is absolutely convergent.
ITo check if the series P
n=1
(1)n
nis absolutely convergent, we need to
check if the series of absolute values P
n=1
1
nis convergent.
ISince P
n=1
1
nis a p-series with p= 1, it diverges and therefore
P
n=1
(1)n
nis not absolutely convergent.
Annette Pilkington Lecture 28 :Absolute Convergence, Ratio and root test
pf3
pf4
pf5
pf8
pf9
pfa

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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