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11.47. Test for convergence the complex power series
     un 1  zn n3 3n 1  n3 jzj jzj    lim  lim Since lim   n!1 n 1 3 3n zn 1  n!1 3 n 1 3 jzj 3 , the series converges for 3 < 1, n!1 un   i.e., jzj < 3, and diverges for jzj > 3. 1 1 X1 X jzjn 1 , so that the series is absolutely For jzj 3, the series of absolute values is 3 3n 1 n n3 n 1 n 1 convergent and thus convergent for jzj 3. Thus, the series converges within and on the circle jzj 3.
zn 1 . n3 3n 1 n 1
11.48. Assuming the power series for ex holds for complex numbers, show that eix cos x i sin x
Letting z ix in ez 1 z z z ; we have 2! 3! ! ! i2 x2 i3 x3 x2 x4 x3 x5 eix 1 ix 1 i x 2! 3! 2! 4! 3! 5! cos x i sin x Similarly, e
ix 2 3
cos x i sin x.
The results are called Euler s identities.
  1 1 1 1 11.49. Prove that lim 1 ln n exists. n!1 2 3 4 n
Letting f x 1=x in (1), Problem 11.11, we nd 1 1 1 1 1 1 1 1 @ ln M @ 1 2 3 4 M 2 3 4 M 1 from which we have on replacing M by n, 1 1 1 1 1 @ 1 ln n @ 1 n 2 3 4 n 1 1 1 1 Thus, the sequence Sn 1 ln n is bounded by 0 and 1. 2 3 4 n
Consider Sn 1 Sn
INFINITE SERIES
[CHAP. 11
  1 n 1 1 1 1 ln . By integrating the inequality @ @ with respect n 1 n n 1 x n to x from n to n 1, we have     1 n 1 1 1 1 1 n 1 @ ln or @ ln @ @0 n 1 n n n 1 n n 1 n i.e., Sn 1 Sn @ 0, so that Sn is monotonic decreasing. Since Sn is bounded and monotonic decreasing, it has a limit. This limit, denoted by
, is equal to 0:577215 . . . and is called Euler s constant. It is not yet known whether
is rational or not.
11.50. Prove that the in nite product
1 1 Y X 1 uk , where uk > 0, converges if uk converges. k 1 k 1
According to the Taylor series for ex (Page 275), 1 x @ ex for x > 0, so that Pn
n Y k 1
1 uk 1 u1 1 u2 1 un @ eu1 eu2 eun eu1 u2 un
Since u1 u2 converges, it follows that Pn is a bounded monotonic increasing sequence and so has a limit, thus proving the required result.
11.51. Prove that the series 1 1 1 1 1 1 is C 1 summable to 1/2.
The sequence of partial sums is 1; 0; 1; 0; 1; 0; . . . . Then S1 1; S1 S2 1 0 1 S1 S2 S3 1 0 1 2 ; ;... : 2 2 3 3 2 3
Continuing in this manner, we obtain the sequence 1; 1 ; 2 ; 1 ; 3 ; 1 ; . . . ; the nth term being 2 3 2 5 2 & 1=2 if n is even Tn . Thus, lim Tn 1 and the required result follows. 2 n!1 n= 2n 1 if n is odd
11.52. (a) If f n 1 x is continuous in a; b prove that for c in a; b , f x f c f 0 c x c 1 00 1 n 1 x 2 n f c x c f c x c x t n f n 1 t dt. 2! n! n! c (b) Obtain the Lagrange and Cauchy forms of the remainder in Taylor s Formula. (See Page 274.)
The proof of (a) is made using mathematical induction. (See 1.) The result holds for n 0 since x f x f c f 0 t dt f c f x f c
We make the induction assumption that it holds for n k and then use integration by parts with dv Then v Thus, 1 k! x
x t k dt and u f k 1 t k!
x t k 1 k 1 !
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