Page 291 - Schaum's Outline of Theory and Problems of Advanced Calculus

282 INFINITE SERIES [CHAP. 11

From Problem 11.11 it follows that
M ð M M 1
X 1 dx X 1
lim < lim < lim
2
2
n þ 1 2 n þ 1
n¼2 n¼1
M!1 M!1 1 x þ 1 M!1
1 1 1
1 1 1
X X X
i.e., < < , from which < as required.
2
2
2
n þ 1 4 n þ 1 4 n þ 1
n¼2 n¼1 n¼1
1 1 1
1 1
X X
Since < ,we obtain, on adding 1 to each side, < þ :
2
2
n þ 1 4 2 n þ 1 2 4
n¼2 n¼1
The required result is therefore proved.
ALTERNATING SERIES
11.15. Given the alternating series a 1 a 2 þ a 3 a 4 þ where 0 @ a nþ1 @ a n and where lim a n ¼ 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
S 2M ¼ða 1 a 2 Þþða 3 a 4 Þþ þ ða 2M 1 a 2M Þ
¼ a 1 ða 2 a 3 Þ ða 4 a 5 Þ ða 2M 2 a 2M 1 Þ a 2M
Since the quantities in parentheses are non-negative, we have
S 2M A 0; S 2 @ S 4 @ S 6 @ S 8 @ @ S 2M @ a 1
Therefore, fS 2M g is a bounded monotonic increasing sequence and thus has limit S.
Also, S 2Mþ1 ¼ S 2M þ a 2Mþ1 . Since lim S 2M ¼ S and lim a 2Mþ1 ¼ 0 (for, by hypothesis,
M!1 M!1
lim a n ¼ 0), it follows that lim S 2Mþ1 ¼ lim S 2M þ lim a 2Mþ1 ¼ S þ 0 ¼ S.
n!1 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
ða 2Mþ1 a 2Mþ2 Þþða 2Mþ3 a 2Mþ4 Þþ ¼ a 2Mþ1 ða 2Mþ2 a 2Mþ3 Þ
and is thus non-negative and less than or equal to a 2Mþ1 ,the first term which is omitted.
Similarly, the error made in stopping after 2M þ 1 terms is
a 2Mþ2 þða 2Mþ3 a 2Mþ4 Þþ ¼ ða 2Mþ2 a 2Mþ3 Þ ða 2Mþ4 a 2Mþ5 Þ
which is non-positive and greater than a 2Mþ2 .

nþ1
1
X ð 1Þ
11.16. (a)Prove that the series converges. (b) Find the maximum error made in approx-
2n 1
n¼1
imating the sum by the first 8 terms and the first 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?
nþ1 1 1
1
1
1
1
(a) The series is 1 þ þ . ð 1Þ .
3 5 7 9 If u n ¼ 2n 1 ,then a n ¼ju n j¼ 2n 1 , a nþ1 ¼ju nþ1 j¼ 2n þ 1
1 1 1
Since @ and since lim ¼ 0, it follows by Problem 11.5(a)that the series
2n þ 1 2n 1 n!1 2n 1
converges.
1 1 1 1 1 1 1 and the
9
3
5
7
(b)Use the results of Problem 11.15(b). Then the first 8 terms give 1 þ þ 11 þ 13 15
error is positive and does not exceed 1 .
17
1 1 1 1 1 1 1 1 and the error is negative and
3 5 7 9 11 13 15 17
Similarly, the first 9 terms are 1 þ þ þ þ
1
greater than or equal to , i.e., the error does not exceed 1 in absolute value.
19 19