Mathematical Methods for Physicists: A concise introduction - Site Map

INFINITE SERIES
Problem A1.11
Show that the error made in stopping after 2M terms is less than or equal to
u 2M‡1 .
Example A1.9
For the series
1 1 2 ‡ 1 3 1 4 ‡ˆX1
nˆ1
…1† n1
;
n
we have u n ˆ…1† n‡1 =n; ju n j ˆ 1=n; ju n‡1 j ˆ 1=…n ‡ 1†. Then for n 1;
ju n‡1 j ju n j. Also we have lim n!1 ju n j ˆ 0. Hence the series converges.
Absolute and conditional convergence
The series P u n is called absolutely convergent if P ju n j converges. If P u n converges
but P ju n j diverges, then P u n is said to be conditionally convergent.
It is easy to show that if P ju n j converges, then P u n converges (in words, an
absolutely convergent series is convergent). To this purpose, let
then
S M ˆ u 1 ‡ u 2 ‡‡u M T M ˆ ju 1 j‡ ju 2 j‡‡ ju M j;
S M ‡ T M ˆ…u 1 ‡ ju 1 j†‡…u 2 ‡ ju 2 j†‡‡…u M ‡ ju M j†
2ju 1 j‡ 2ju 2 j‡‡2ju M j:
Since P ju n j converges and since u n ‡ ju n j 0, for n ˆ 1; 2; 3; ...; it follows that
S M ‡ T M is a bounded monotonic increasing sequence, and so
lim M!1 …S M ‡ T M † exists. Also lim M!1 T M exists (since the series is absolutely
convergent by hypothesis),
lim S M ˆ lim …S M ‡ T M T M †ˆ lim …S M ‡ T M † lim T M
M!1 M!1 M!1 M!1
must also exist and so the series P u n converges.
The terms of an absolutely convergent series can be rearranged in any order,
and all such rearranged series will converge to the same sum. We refer the reader
to text-books on advanced calculus for proof.
Problem A1.12
Prove that the series
converges.
1 1 2 2 ‡ 1 3 2 1 4 2 ‡ 1 5 2
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