Telescoping Sums. Dr. Philippe B. Laval Kennesaw State University. Abstract

Transcription

1 Telescopng Sums Dr. Phlppe B. Laval Kennesaw State Unversty Abstract Ths hand out s a descrpton of the technque known as telescopng sums, whch s used when studyng the convergence of some seres. Telescopng sums. Introducton The technque we are about to descrbe apples to seres of the form ) where p s some postve nteger. It s used when we are studyng the convergence oftheseseres. Forexample,wewouldusettostudytheseres +), +2),,... It nvolves two steps. Frst, we wrte the general +3) term of the seres as a dfference of two fractons, usng partal fracton decomposton. Second, we fnd and smplfy the sequence of partal sums, as most of ts terms wll cancel. 2 Partal Fracton Decomposton. Theorem ) = p p ) = p ) Proof. Ths s easly done usng the technques of partal fracton decomposton. Frst, we notce that ) = A + B ) If we multply each sde by the denomnator of the fracton on the left, we obtan =A )+B

2 When =0, we obtan Ap = A = p When = p, we obtan Bp = B = p Replacng those values n equaton gves the desred result. Example 2 If p =, we have +) = + Example 3 If p =2, we have +2) = 2 ) +2 Example 4 If p =3, we have +3) = 3 ) +3 3 Smplfyng the Sequence of Partal Sums Instead of studyng, we study ) p ). In fact, because we wll prove that ) converges, t s enough to study ) snce We now concentrate on p ) = p partal sums, S n ). By defnton, ) 2) ). Tostudyt,weusetssequenceof ) We frst look at the specal cases p =and p =2. We then generalze our result for any postve nteger p. 2

3 3. Case p = In ths case, ) + Thekeyherestonotcethat and wll generate the same values wth + opposte sgns. Hence, they wll cancel. Of course, these same values wll be generated for dfferent values of. More precsely, they wll generate the same values for values of whch are unt apart. For a certan value of, generates the same value that when =4, generated for the prevous value of. For example, + = 4 whch s the same value that gves when =3. + Ths means that everythng wll cancel except the value generated by for the startng value of, and the last value of for the endng value of. When + =, =. =when =0,but s never 0. Smlarly, when = n, + + = n + whch can only be generated by for = n +.Butsnever n +. Therefore, everythng cancels except for the startng value of and for the endng value of. Therefore, we see that + n + Ths can be verfed f we expand S n Snce It follows that n n + n n + = n + lm n ) = Case p =2 In ths case, ) +2 3

4 Ths s smlar to above, but ths tme, and wll generate the same +2 values wth opposte sgns for values of whch are 2 unts apart. Therefore, usng the same reasonng as above, we see that the frst two values generated by and the last two values generated by wll not cancel. Everythng else +2 wll cancel. Hence, + 2 n + n +2 We see that lm + n 2 Therefore ) = General case In ths case, ) Ths s smlar to above, but ths tme, and wll generate the same values wth opposte sgns for values of whch are p unts apart. Therefore, usng the same reasonng as above, we see that the frst p values generated by and the last p values generated by wll not cancel. Everythng else wll +2 cancel. Hence, p n + n n + p We see that Therefore lm n p ) = p p = 4

5 3.4 Concluson The above has shown us that f p s any postve nteger, then ) p = 3) Our orgnal problem was to study Snce. We saw above that ) ) = p ) ) converges, we have by equaton 2 ) = p ) Usng equaton 3, we obtan the followng theorem Theorem 5 If p s a postve nteger, then ) = p p 4) Corollary 6 If p s a postve nteger, and C s a constant, then C ) = C p p 5) Example 7 Study the convergence of +) Ths s a seres lke the one n equaton 4 wth p =. Therefore, +) = = You wll recall that we had already derved ths result as an example n the handout on seres. 5

6 Example 8 Study the convergence of +2) Ths s a seres lke the one n equaton 4 wth p =2. Therefore, +2) = 2 = ) 2 = 3 4 You wll recall that we had already found ths result as part as one of the exercses assgned. 5 Example 9 Study the convergence of +3) Ths s a seres lke the one n equaton 5 wth p =3and C =5. Therefore, 5 +3) = 5 3 = ) 3 =