1550251515-Classical_Complex_Analysis__Gonzalez_

196 Chapter4

so that I: lanl converges if f3 > 1 and diverges if f3 < 1. For the case (3 = 1,
we have, using (f),

lim lnn [n (-

1

lanl I -1)-1] = lim (lnn)O(ni--y) = 0

n->oo an+i n->oo

which shows that I: Ian I also diverges in this case.

(h) Gauss test. If

~ _ nP + ainp-i + · · · + ap

an+i nP + f3inp-i + · · · + (3p

where p and the coefficients aj, /3j (j = 1, ... ,p) are independent of n, then

I: lanl converges if Re(ai - /3i) > 1, and diverges if Re(ai - /3i)::::; 1.

Proof Letting ai = Ai+ iA 2 , /3i = Bi+ iB2, we have

--an = 1 + ( ai - /3i ) - +^1 0( n -2)

an+i n

=l+ Ai-Bi +iA2-B2 +O(n-2)

n n

Hence

Ai - Bi ·

=l+ +O(n-^2 )

n

assuming n large enough so that 1 +(Ai - Bi)/n > 0. The conclusion

now follows from test (g).

Note Throughout the preceding computation the symbol O(n-^2 ) does

not always stand for the same function of n.

Example Consider the series

1

ab a(a + 1)

+ l!c + 2!c(c+l) +···

+ a(a+l)···(a+n-l)b(b+l)···(b+n-1) +···

n!c( c + 1) · · · ( c + n - 1) ·

where the complex constants a, b, c are neither zero nor negative integers.

Here we have

an =~-~-~ ( n + 1 )( c + n)

an+l (a+ n)(b + n)

n^2 + ( c + 1 )n + c

n2+(a+b)n+ab