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term of a series does not approach zero the series is divergent. 2. 3. Multiplication of each term of a series by a constant di erent from zero does not a ect the convergence or divergence. Removal (or addition) of a nite number of terms from (or to) a series does not a ect the convergence or divergence.
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SPECIAL SERIES 1. Geometric series S
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arn 1 a ar ar2 , where a and r are constants, converges to
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n 1 a a 1 rn if jrj < 1 and diverges if jrj A 1. The sum of the rst n terms is Sn 1 r 1 r (see Problem 2.25, Chap. 2). 1 X 1 1 1 1 The p series ; where p is a constant, converges for p > 1 and diverges n p 1p 2p 3p n 1
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for p @ 1.
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The series with p 1 is called the harmonic series.
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TESTS FOR CONVERGENCE AND DIVERGENCE OF SERIES OF CONSTANTS More often than not, exact values of in nite series cannot be obtained. Thus, the search turns toward information about the series. In particular, its convergence or divergence comes in question. The following tests aid in discovering this information.
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CHAP. 11]
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INFINITE SERIES
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1. Comparison test for series of non-negative terms. (a) Convergence. Let vn A 0 for all n > N and suppose that vn converges. Then if 0 @ un @ vn for all n > N, un also converges. Note that n > N means from some term onward. Often, N 1.
EXAMPLE: Since X 1 X 1 1 1 converges, @ n and also converges. 2n 1 2 2n 2n 1
(b) Divergence. Let vn A 0 for all n > N and suppose that vn diverges. Then if un A vn for all n > N, un also diverges.
EXAMPLE: Since
1 1 X1 X 1 1 1 > and diverges, also diverges. ln n n n ln n n 2 n 2
2. The Limit-Comparison or Quotient Test for series of non-negative terms. u (a) If un A 0 and vn A 0 and if lim n A 6 0 or 1, then un and vn either both converge n!1 vn or both diverge. (b) If A 0 in (a) and vn converges, then un converges. (c) If A 1 in (a) and vn diverges, then un diverges. This test is related to the comparison test and is often a very useful alternative to it. particlar, taking vn 1=np , we have from known facts about the p series the Theorem 1. Let lim np un A. Then
(i) un converges if p > 1 and A is nite. (ii) un diverges if p @ 1 and A 6 0 (A may be in nite).
EXAMPLES: 1: 2: X n converges since 4n3 2
lim n2
n 1 : 4n3 2 4 ln n 1: n 1 1=2
X ln n p diverges since n 1
lim n1=2
3. Integral test for series of non-negative terms. If f x is positive, continuous, and monotonic decreasing for x A N and is such that f n un ; n N; N 1; N 2; . . . , then un converges or diverges according as
f x dx lim
f x dx converges or diverges.
In particular we may have N 1, as
is often true in practice. This theorem borrows from the next chapter since the integral has an unbounded upper limit. (It is an improper integral. The convergence or divergence of these integrals is de ned in much the same way as for in nite series.)
EXAMPLE:
1 X 1 converges since n2 n 1
  dx 1 lim 1 exists. M x2 M!1
4. Alternating series test. An alternating series is one whose successive terms are alternately positive and negative. An alternating series converges if the following two conditions are satis ed (see Problem 11.15). (a) jun 1 j @ jun j for n A N (Since a xed number of terms does not a ect the convergence or divergence of a series, N may be any positive integer. Frequently it is chosen to be 1.)   (b) lim un 0 or lim jun j 0
n!1 n!1
EXAMPLE. jun 1 j
INFINITE SERIES
[CHAP. 11
For the series 1 1 1 1 1 2 3 4 5 Then for n A 1, jun 1 j @ jun j.
1 . n 1
1 X 1 n 1 1 n 1 1 , we have un , jun j , n n n n 1
Also lim jun j 0.
Hence, the series converges.
Theorem 2. The numerical error made in stopping at any particular term of a convergent alternating series which satis es conditions (a) and (b) is less than the absolute value of the next term.
EXAMPLE. 1 5 0:2. If we stop at the 4th term of the series 1 1 1 1 1 , the error made is less than 2 3 4 5
Absolute and conditional convergence. The series un is called absolutely convergent if jun j converges. If un converges but jun j diverges, then un is called conditionally convergent. In words, an absolutely convergent series is
Theorem 3. If jun j converges, then un converges. convergent (see Problem 11.17).
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