PureMath.pdf

# And 1 1 2 2 1 3 2 1 5 2 1 6 2 1 7 2 1 9 2 15 16 1 1 2

• 587

This preview shows pages 403–406. Sign up to view the full content.

and 1 + 1 2 2 + 1 3 2 + 1 5 2 + 1 6 2 + 1 7 2 + 1 9 2 + · · · = 15 16 1 + 1 2 2 + 1 3 2 + . . . . [To prove the first result we note that 1 + 1 2 2 + 1 3 2 + . . . = 1 + 1 2 2 + 1 3 2 + 1 4 2 + . . . = 1 + 1 3 2 + 1 5 2 + · · · + 1 2 2 1 + 1 2 2 + 1 3 2 + . . . , by theorems (8) and (6) of § 77 .] 14. Prove by a reductio ad absurdum that (1 /n ) is divergent. [If the series were convergent we should have, by the argument used in Ex. 13, 1 + 1 2 + 1 3 + · · · = (1 + 1 3 + 1 5 + . . . ) + 1 2 (1 + 1 2 + 1 3 + . . . ) , or 1 2 + 1 4 + 1 6 + · · · = 1 + 1 3 + 1 5 + . . . which is obviously absurd, since every term of the first series is less than the corresponding term of the second.]

This preview has intentionally blurred sections. Sign up to view the full version.

[VIII : 170] THE CONVERGENCE OF INFINITE SERIES, ETC. 388 169. Before proceeding further in the investigation of tests of con- vergence and divergence, we shall prove an important general theorem concerning series of positive terms. Dirichlet’s Theorem. * The sum of a series of positive terms is the same in whatever order the terms are taken. This theorem asserts that if we have a convergent series of positive terms, u 0 + u 1 + u 2 + . . . say, and form any other series v 0 + v 1 + v 2 + . . . out of the same terms, by taking them in any new order, then the second series is convergent and has the same sum as the first. Of course no terms must be omitted: every u must come somewhere among the v 0 s, and vice versa . The proof is extremely simple. Let s be the sum of the series of u 0 s. Then the sum of any number of terms, selected from the u 0 s, is not greater than s . But every v is a u , and therefore the sum of any number of terms selected from the v 0 s is not greater than s . Hence v n is convergent, and its sum t is not greater than s . But we can show in exactly the same way that s 5 t . Thus s = t . 170. Multiplication of Series of Positive Terms. An immediate corollary from Dirichlet’s Theorem is the following theorem: if u 0 + u 1 + u 2 + . . . and v 0 + v 1 + v 2 + . . . are two convergent series of positive terms, and s and t are their respective sums, then the series u 0 v 0 + ( u 1 v 0 + u 0 v 1 ) + ( u 2 v 0 + u 1 v 1 + u 0 v 2 ) + . . . is convergent and has the sum st . * This theorem seems to have first been stated explicitly by Dirichlet in 1837. It was no doubt known to earlier writers, and in particular to Cauchy.
[VIII : 170] THE CONVERGENCE OF INFINITE SERIES, ETC. 389 Arrange all the possible products of pairs u m v n in the form of a doubly infinite array u 0 v 0 u 1 v 0 u 2 v 0 u 3 v 0 . . . u 0 v 1 u 1 v 1 u 2 v 1 u 3 v 1 . . . u 0 v 2 u 1 v 2 u 2 v 2 u 3 v 2 . . . u 0 v 3 u 1 v 3 u 2 v 3 u 3 v 3 . . . . . . . . . . . . . . . . . . . We can rearrange these terms in the form of a simply infinite series in a variety of ways. Among these are the following. (1) We begin with the single term u 0 v 0 for which m + n = 0; then we take the two terms u 1 v 0 , u 0 v 1 for which m + n = 1; then the three terms u 2 v 0 , u 1 v 1 , u 0 v 2 for which m + n = 2; and so on. We thus obtain the series u 0 v 0 + ( u 1 v 0 + u 0 v 1 ) + ( u 2 v 0 + u 1 v 1 + u 0 v 2 ) + . . . of the theorem. (2) We begin with the single term u 0 v 0 for which both suffixes are zero; then we take the terms u 1 v 0 , u 1 v 1 , u 0 v 1 which involve a suffix 1 but no higher suffix; then the terms u 2 v 0 , u 2 v 1 , u 2 v 2 , u 1 v 2 , u 0 v 2 which involve a suffix 2 but no higher suffix; and so on. The sums of these groups of terms are respectively equal to u 0 v 0 , ( u 0 + u 1 )(

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.
• Fall '14

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern