Summing from n 1 to M 1, u2 u3 uM @ M in .NET framework

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Summing from n 1 to M 1, u2 u3 uM @ M
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f x dx @ u1 u2 uM 1
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If f x is strictly decreasing, the equality signs in (1) can be omitted.
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CHAP. 11] M If
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INFINITE SERIES
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f x dx exists and is equal to S, we see from the left-hand inequality in (1) that
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u2 u3 uM is monotonic increasing and bounded above by S, so that un converges. M f x dx is unbounded, we see from the right-hand inequality in (1) that un diverges. If lim
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Thus the proof is complete.
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11.12. Illustrate geometrically the proof in Problem 11.11.
Geometrically, u2 u3 uM is the total area of the rectangles shown shaded in Fig. 11-3, while u1 u2 uM 1 is the total area of the rectangles which are shaded and nonshaded. The area under the curve y f x from x 1 to x M is intermediate in value between the two areas given above, thus illustrating the result (1) of Problem 11.11.
1 X 1 ; p constant; nP 1 1 1 X n2 X 1 ; d ne . c n ln n 2 1
Fig. 11-3
11.13. Test for convergence: b
1 X 1
n ; n2 1
Consider
dx xp
x p dx
 x1 p M M 1 p 1  where p 6 1: 1 p 1 1 p
If p < 1; lim
M 1 p 1 1, so that the integral and thus the series diverges. M!1 1 p M 1 p 1 1 , so that the integral and thus the series converges. 1 p p 1
If p > 1; lim
M M dx dx ln M and lim ln M 1, so that the integral and thus the series If p 1, p M!1 1 x 1 x diverges. Thus, the series converges if p > 1 and diverges if p @ 1. b M
M!1 1
x dx lim 1 ln x2 1 jM lim 1 ln M 2 1 1 ln 2 1 and the series diverges. 1 2 2 1 M!1 2 M!1 2 x dx lim ln ln x jM lim fln ln M ln ln 2 g 1 and the series diverges. 2 M!1 x ln x M!1 xe x dx lim 1 e x jM lim 1 2
M!1 2
M!1 1
1 1 e M!1 2
1 e M 2
1 e 1 and the series converges. 2
Note that when the series converges, the value of the corresponding integral is not (in general) the same as the sum of the series. However, the approximate sum of a series can often be obtained quite accurately by using integrals. See Problem 11.74.
1  X 1 1  < < . 4 n 1 n2 1 2 4
11.14. Prove that
INFINITE SERIES
[CHAP. 11
From Problem 11.11 it follows that 1 < lim M!1 n2 1 M!1 n 2 lim i.e.,
1 X M X
M 1 X 1 dx < lim x2 1 M!1 n 1 n2 1
1 1 1  X 1  X 1 < < < , from which as required. 4 n 1 n2 1 n2 1 4 n 1 n2 1 n 2 1 X
Since
1  < , we obtain, on adding n2 1 4 n 2
to each side,
1 1  < : n2 1 2 4 n 1
The required result is therefore proved.
ALTERNATING SERIES 11.15. Given the alternating series a1 a2 a3 a4 where 0 @ an 1 @ an and where lim an 0. n!1 Prove that (a) the series converges, (b) the error made in stopping at any term is not greater than the absolute value of the next term.
(a) The sum of the series to 2M terms is S2M a1 a2 a3 a4 a2M 1 a2M a1 a2 a3 a4 a5 a2M 2 a2M 1 a2M Since the quantities in parentheses are non-negative, we have S2M A 0; S2 @ S4 @ S6 @ S8 @ @ S2M @ a1 Therefore, fS2M g is a bounded monotonic increasing sequence and thus has limit S. Also, S2M 1 S2M a2M 1 . Since lim S2M S and lim a2M 1 0 (for, by hypothesis,
M!1 M!1 n!1
lim an 0), it follows that lim S2M 1 lim S2M lim a2M 1 S 0 S.
M!1 M!1 M!1
Thus, the partial sums of the series approach the limit S and the series converges. (b) The error made in stopping after 2M terms is a2M 1 a2M 2 a2M 3 a2M 4 a2M 1 a2M 2 a2M 3 and is thus non-negative and less than or equal to a2M 1 , the rst term which is omitted. Similarly, the error made in stopping after 2M 1 terms is a2M 2 a2M 3 a2M 4 a2M 2 a2M 3 a2M 4 a2M 5 which is non-positive and greater than a2M 2 .
1 X 1 n 1 converges. (b) Find the maximum error made in approx2n 1 n 1 imating the sum by the rst 8 terms and the rst 9 terms of the series. (c) How many terms of the series are needed in order to obtain an error which does not exceed .001 in absolute value
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