PureMath.pdf

# V v 1 u v u u 1 u 2 v v 1 v 2 u u 1 v v 1 and the sum

This preview shows pages 405–408. Sign up to view the full content.

v 0 + v 1 ) - u 0 v 0 , ( u 0 + u 1 + u 2 )( v 0 + v 1 + v 2 ) - ( u 0 + u 1 )( v 0 + v 1 ) , . . . and the sum of the first n + 1 groups is ( u 0 + u 1 + · · · + u n )( v 0 + v 1 + · · · + v n ) , and tends to st as n → ∞ . When the sum of the series is formed in this manner the sum of the first one, two, three, . . . groups comprises all the terms in the first, second, third, . . . rectangles indicated in the diagram above. The sum of the series formed in the second manner is st . But the first series is (when the brackets are removed) a rearrangement of the second; and therefore, by Dirichlet’s Theorem, it converges to the sum st . Thus the theorem is proved.

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

View Full Document
[VIII : 171] THE CONVERGENCE OF INFINITE SERIES, ETC. 390 Examples LXVIII. 1. Verify that if r < 1 then 1 + r 2 + r + r 4 + r 6 + r 3 + · · · = 1 + r + r 3 + r 2 + r 5 + r 7 + · · · = 1 / (1 - r ) . 2. * If either of the series u 0 + u 1 + . . . , v 0 + v 1 + . . . is divergent, then so is 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 ) + . . . , except in the trivial case in which every term of one series is zero. 3. If the series u 0 + u 1 + . . . , v 0 + v 1 + . . . , w 0 + w 1 + . . . converge to sums r , s , t , then the series λ k , where λ k = u m v n w p , the summation being extended to all sets of values of m , n , p such that m + n + p = k , converges to the sum rst . 4. If u n and v n converge to sums s and t , then the series w n , where w n = u l v m , the summation extending to all pairs l , m for which lm = n , converges to the sum st . 171. Further tests for convergence and divergence. The exam- ples on pp. 385 387 suffice to show that there are simple and interesting types of series of positive terms which cannot be dealt with by the general tests of § 168 . In fact, if we consider the simplest type of series, in which u n +1 /u n tends to a limit as n → ∞ , the tests of § 168 generally fail when this limit is 1. Thus in Ex. lxvii . 5 these tests failed, and we had to fall back upon a special device, which was in essence that of using the series of Ex. lxvii . 4 as our comparison series, instead of the geometric series. The fact is that the geometric series, by comparison with which the tests of § 168 were obtained, is not only convergent but very rapidly convergent, far more rapidly than is necessary in order to ensure convergence. The tests derived from comparison with it are therefore naturally very crude, and much more delicate tests are often wanted. We proved in Ex. xxvii . 7 that n k r n 0 as n → ∞ , provided r < 1, whatever value k may have; and in Ex. lxvii . 1 we proved more than this, viz. that the series n k r n is convergent. It follows that the sequence r , r 2 , r 3 , . . . , r n , . . . , where r < 1, diminishes more rapidly than the sequence 1 - k , 2 - k , 3 - k , . . . , n - k , . . . . This seems at first paradoxical if r is not much less than unity, and k is large. Thus of the two sequences 2 3 , 4 9 , 8 27 , . . . ; 1 , 1 4096 , 1 531 , 441 , . . .
[VIII : 172] THE CONVERGENCE OF INFINITE SERIES, ETC. 391 whose general terms are ( 2 3 ) n and n - 12 , the second seems at first sight to decrease far more rapidly. But this is far from being the case; if only we go far enough into the sequences we shall find the terms of the first sequence very much the smaller. For example, (2 / 3) 4 = 16 / 81 < 1 / 5 , (2 / 3) 12 < (1 / 5) 3 < (1 / 10) 2 , (2 / 3) 1000 < (1 / 10) 166 , while 1000 - 12 = 10 - 36 ;

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ 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