1550251515-Classical_Complex_Analysis__Gonzalez_

624 Chapter 8

If in (8.20-20) we let n = 1, 2, 3, ... , n - 1 and add the resulting

inequalities, we obtain

[ ( )

2 3 2 n n-1 l ( ) 1 2 3 n n-1

ln - - · · · (--) + -ln - - · · · -- = n - 1 + L ak

1 2 n - 1 2 1 2 n - 1 k=l

or

n 1 n-1

· ln !!:__ n! + - 2 ln n = n - 1 + " L,,; ak

k=l

Similarly, from (8.20-21) we get

n-l 1 ( 1) 1

"ak L,,; < - 12 1 - -n · < - 12

k=l

(8.20-22)

(8.20-23)

Therefore, if we take antilogarithms in (8.20-22) and make use of (8.20-23),

we find that

(8.20-24)

or

__ ennl ._ - 1-~n-1 L....1 Olk 11/12

nn+1/2 - e > e (8.20-25)

The sequence on the left of (8.20-25) is strictly decreasing (as the equality

shows), and bounded below by e^11112. Hence it has a positive limit, say

L. We have

e2n(n1)2 (2n)2nH/2

L = I}L-^1 = lim. · lim

n--+oo n2n+l n--+oo e2n(2n )!

. (n!)222n+l/2
= J~1!, (2n)!nl/2 (8.20-26)

The so-called Vvallis formula, namely,

II

"" ( 2n 2n ) 1

n=l 2n-1. 2n+l = 211"

can also be written as

. (n!)2. 22n
hm = ,,/1r
n--+oo (2n )!nl/2

I.

which shows that the limit sought in (8.20-26) is L = V2if.
Hence

• enn!.
  n-+oo bm -+1/2 nn =...&