1550251515-Classical_Complex_Analysis__Gonzalez_

Sequences and Series

taking Dn = nlnn in (c). Then

Dn+1 = (n + 1) ln(n + 1) = (n + 1) (inn+ ln ( 1 + ~)]

=(n+l) [inn+~ +O(n-^2 )]

1

=(n+l)lnn+l+ - +O(n-^1 )

n

so that

Dn-lanl lanl ( -1

1

--

1

• Dn+1 = nlnn-
  1

--

1

• n + l)lnn -1-O(n )
  an+1 an+1

Hence

= lnn [n (~ -1)-1] -1-O(n-^1 )

lan+1I

lim (nn -

1

lanl I - Dn+ 1 ) = lim Inn [n (-

1

lanl I -1) -1] -1

n-+oo an+1 n->oo an+1

and it follows that if

lim lnn[n(-

1

1anl

1

-1)-1] >1

n-+oo an+1

the series 2= Ian I converges, and if

lim ln n [n (-

1

Ian I I - 1) -1] < 1

n-+oo an+1

the series diverges.

195

There are also a number of special rules discussed in treatises on series,
some of which can be obtained from the preceding tests. As an example,
we prove the following:
(g) Suppose that

~ =a:+ f!_ + O(n-,.,,)

lan+1 I n

where a:, (:J, 1 are real constants with 'Y > 1. Then, if a: > 1, the series

2= Ian I converges, and if a: < 1, 2= Ian I diverges. If a: = 1, then 2= Ian I

converges if (:J > 1 and diverges if (:J :::; 1.

Proof The first part follows at once from the ratio test, since lim(lanl/
lan+1 I) = a:. For case a: = 1 we have, using Raabe's test,

hm. n ( --Ian! -1 ) = (:J

n-+oo lan+1 I