1550251515-Classical_Complex_Analysis__Gonzalez_

Sequences and Series 207

superior and L2 the limit inferior of lan+1l/lanl, as stated in the following

theorem, due to S. Pincherle.

Theorem 4.21 Let R be the radius of convergence of I: anzn (an -=/= Q),

and suppose that

1. 1msup I --an+1 I = L 1

n-+oo an

and 1.. 1mm f I --an+1 I = L 2

n--+oo an

Then l/L1 ::::; R ::::; 1/L 2.

Proof The third form of the ratio test gives

hmsup. I an+1zn+1 I = Lilzl < 1

n-+oo anzn

for convergence, and

or

or

1

lzl < Li

1

lzl > L2

for divergence. Hence the power series does not diverge if lzl < 1/ L 1 , and

it does not converge if lzl > l/L 2 • Thus, 1/L 1 ::::; R::::; 1/L 2 •

In some cases it is possible to give a precise formula for the computation
of R, as in the next theorem (Gonzalez [4]).

Theorem 4.22 Suppose that the ordinary limits

1. lm I akm+r I = L r

m-+oo ak(m+l)+r

( k > 1 fixed, r = 0, ... , k - 1)

exist, and let L = min(L 0 , ••. , Lk-d· Then the radius of convergence of
I: anzn is given by

(4.11-7)

Alternatively,

R =^1 1m.. 1n f k I --an I

n-+oo an+k

(4.11-8)

Proof It suffices to decompose I: anzn into k power series of the form
00
~ ~ akm+rZ· km+r (r = 0,1, ... ,k-1)
m=O

For the rth series we have, by the ratio test,