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[Hint: Use the Cauchy convergence criterion, Page 25.] 11.120. Prove that the hypergeometric series (Page 276) (a) is absolutely convergent for jxj < 1, (b) is divergent for jxj > 1, (c) is absolutely divergent for jxj 1 if a b c < 0; d satis es the di erential equation x 1 x y 00 fc a b 1 xgy 0 aby 0. 11.121. If F a; b; c; x is the hypergeometric function de ned by the series on Page 276, prove that (a) F p; 1; 1; x 1 x p ; b xF 1; 1; 2; x ln 1 x ; c F 1 ; 1 ; 3 ; x2 sin 1 x =x. 2 2 2 11.122. Find the sum of the series S x x x3 x5 . 1 3 1 3 5 0 [Hint: Show that S x 1 xS x and solve.] x 2 2 Ans: ex =2 e x =2 dx
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11.123. Prove that 1
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11.124. Establish the Dirichlet test on Page 270.
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11.125. Prove that
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is uniformly convergent in any interval which does not include 0; ; 2; . . . .
[Hint: use the Dirichlet test, Page 270, and Problem 1.94, 1.] 11.126. Establish the results on Page 275 concerning the binomial series. [Hint: Examine the Lagrange and Cauchy forms of the remainder in Taylor s theorem.]
1 X 1 n 1 n 1
INFINITE SERIES
[CHAP. 11
11.127. Prove that
n x2
converges uniformly for all x, but not absolutely.
1 1 1  1 11.128. Prove that 1 p ln 2 4 7 10 3 3 3 11.129. If x ye y , prove that y
1 X 1 n 1 nn 1 n 1
xn for 1=e < x @ 1=e.
11.130. Prove that the equation e   1 has only one real root and show that it is given by  1
1 X 1 n 1 nn 1 e n n 1
11.131. Let
x B x2 B x3 1 B1 x 2 3 . (a) Show that the numbers Bn , called the Bernoulli numbers, 1 2! 3! satisfy the recursion formula B 1 n Bn 0 where B k is formally replaced by Bk after expanding. (b) Using (a) or otherwise, determine B1 ; . . . ; B6 . ex
1 1 b B1 1 ; B2 1 ; B3 0; B4 30 ; B5 0; B6 42. 2 6
Ans:
11.132. (a) Prove that k 1; 2; 3; . . . :
 x x x coth 1 : e 1 2 2
b Use Problem 11.127 and part (a) to show that B2k 1 0 if
11.133. Derive the series expansions: a coth x 1 x x3 B 2x 2n 2n x 3 45 2n !x
b cot x
1 x x3 B 2x 2n 1 n 2n x 3 45 2n !x x3 2x5 2 22n 1 B2n 2x 2n 1 1 n 1 3 15 2n !
c tan x x
d csc x
1 x 7 3 2 22n 1 1 B2n x2n 1 x 1 n 1 x 6 360 2n !
[Hint: For (a) use Problem 11.132; for (b) replace x by ix in (a); for (c) use tan x cot x 2 cot 2x; for (d) use csc x cot x tan x=2.] 11.134. Prove that
1 Y n 1
1 n3
 converges.  1 Y 1 diverges. 1 n n 1
11.135. Use the de nition to prove that
11.136. Prove that
1 Y 1 un , where 0 < un < 1, converges if and only if un converges. n 1
11.137. (a) Prove that
 1 Y 1 1 2 converges to 1. (b) Evaluate the in nite product in (a) to 2 decimal places and 2 n n 2 compare with the true value.
11.138. Prove that the series 1 0 1 1 0 1 1 0 1 is the C 1 summable to zero.
CHAP. 11]
INFINITE SERIES
11.139. Prove that the Cesaro method of summability is regular.
[Hint: See Page 278.]
11.140. Prove that the series 1 2x 3x2 4x3 nxn 1 converges to 1= 1 x 2 for jxj < 1. 11.141. A series a
1 X n 0 1 X 1 n n 1 is Abel summable to 1/4 and n 0 1 X 1 n n 1 n 2 n 0
an is called Abel summable to S if S lim
1 X n 0
x!1
an xn exists.
Prove that
is Abel summable to 1/8.
1 1 XX
11.142. Prove that the double series p > 1 or p @ 1, respectively. 1 11.143. (a) Prove that
1 , where p is a constant, converges or diverges according as m2 n2 p m 1 n 1 1
ex u 1 1 2! 3! 1 n 1 n 1 ! 1 n n! du 2 3 4 x x xn u x x x 1 x u e 1 1 2! 3! du \$ 2 3 4 b Use a to prove that u x x x x x
ex u du. un 1
Improper Integrals
DEFINITION OF AN IMPROPER INTEGRAL The functions that generate the Riemann integrals of 6 are continuous on closed intervals. Thus, the functions are bounded and the intervals are nite. Integrals of functions with these characteristics are called proper integrals. When one or more of these restrictions is relaxed, the integrals are said to be improper. Categories of improper integrals are established below. b The integral f x dx is called an improper integral if
1. 2.
a 1 or b 1 or both, i.e., one or both integration limits is in nite, f x is unbounded at one or more points of a @ x @ b. Such points are called singularities of f x .
Integrals corresponding to (1) and (2) are called improper integrals of the rst and second kinds, respectively. Integrals with both conditions (1) and (2) are called improper integrals of the third kind.
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