If z ρ we obtain a similar expansion in descending

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If | z | > ρ we obtain a similar expansion in descending powers, while if | z | = ρ no such expansion is possible. 20. Show that if | z | < 1 then 1 + 2 z + 3 z 2 + · · · + ( n + 1) z n + · · · = 1 / (1 - z ) 2 .
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[IV : 88] LIMITS OF FUNCTIONS OF A 194 [The sum to n terms is 1 - z n (1 - z ) 2 - nz n 1 - z .] 21. Expand L/ ( z - α ) 2 in powers of z , ascending or descending according as | z | < | α | or | z | > | α | . 22. Show that if b 2 = ac and | az | < | b | then Az + B az 2 + 2 bz + c = X 0 p n z n , where p n = { ( - a ) n /b n +2 }{ ( n + 1) aB - nbA } ; and find the corresponding ex- pansion, in descending powers of z , which holds when | az | > | b | . 23. Verify the result of Ex. 19 in the case of the fraction 1 / (1 + z 2 ). [We have 1 / (1 + z 2 ) = z n sin { 1 2 ( n + 1) π } = 1 - z 2 + z 4 - . . . .] 24. Prove that if | z | < 1 then 1 1 + z + z 2 = 2 3 X 0 z n sin { 2 3 ( n + 1) π } . 25. Expand (1 + z ) / (1 + z 2 ), (1 + z 2 ) / (1 + z 3 ) and (1 + z + z 2 ) / (1 + z 4 ) in ascending powers of z . For what values of z do your results hold? 26. If a/ ( a + bz + cz 2 ) = 1 + p 1 z + p 2 z 2 + . . . then 1 + p 2 1 z + p 2 2 z 2 + · · · = a + cz a - cz a 2 a 2 - ( b 2 - 2 ac ) z + c 2 z 2 . ( Math. Trip. 1900.) 27. If lim n →∞ s n = l then lim n →∞ s 1 + s 2 + · · · + s n n = l. [Let s n = l + t n . Then we have to prove that ( t 1 + t 2 + · · · + t n ) /n tends to zero if t n does so. We divide the numbers t 1 , t 2 , . . . , t n into two sets t 1 , t 2 , . . . , t p and t p +1 , t p +2 , . . . , t n . Here we suppose that p is a function of n which tends to as n → ∞ , but more slowly than n , so that p → ∞ and p/n 0: e.g. we might suppose p to be the integral part of n .
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[IV : 88] POSITIVE INTEGRAL VARIABLE 195 Let be any positive number. However small may be, we can choose n 0 so that t p +1 , t p +2 , . . . , t n are all numerically less than 1 2 when n = n 0 , and so | ( t p +1 + t p +2 + · · · + t n ) /n | < 1 2 ( n - p ) /n < 1 2 . But, if A is the greatest of the moduli of all the numbers t 1 , t 2 , . . . , we have | ( t 1 + t 2 + · · · + t p ) /n | < pA/n, and this also will be less than 1 2 when n = n 0 , if n 0 is large enough, since p/n 0 as n → ∞ . Thus | ( t 1 + t 2 + · · · + t n ) /n | 5 | ( t 1 + t 2 + · · · + t p ) /n | + | ( t p +1 + · · · + t n ) /n | < when n = n 0 ; which proves the theorem. The reader, if he desires to become expert in dealing with questions about limits, should study the argument above with great care. It is very often neces- sary, in proving the limit of some given expression to be zero, to split it into two parts which have to be proved to have the limit zero in slightly different ways. When this is the case the proof is never very easy. The point of the proof is this: we have to prove that ( t 1 + t 2 + · · · + t n ) /n is small when n is large, the t ’s being small when their suffixes are large. We split up the terms in the bracket into two groups. The terms in the first group are not all small, but their number is small compared with n . The number in the second group is not small compared with n , but the terms are all small, and their number at any rate less than n , so that their sum is small compared with n . Hence each of the parts into which ( t 1 + t 2 + · · · + t n ) /n has been divided is small when n is large.] 28. If φ ( n ) - φ ( n - 1) l as n → ∞ , then φ ( n ) /n l .
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