ln(l+x ) =x-

-x2

+ R3{x),

where Rs(x) is a remainder term satisfying

\R3(x)\ = C\x\\ |z| 1/2.

(You may wish to review Taylor's theorem with remainder to find

such a C.) Mathematicians often write this as

ln(l + x)=x-

-x2

+ 0(x

3

),

where

0(x3)

represents a function that is bounded by a constant times

x3. In particular, for all x small enough

|m(l + x)| 2\x\.

In particular, if the {aj} is an absolutely convergent series, then the

limit

lim lnTT[l + a,j]

3 = 1

exists.

Returning to the problem at hand, let

J

To see how close this is to 1, we again take logs and expand in the

Taylor series,

In

r ii

i + T

L

J\

j

= j l n

r ii

l + T

L

J\

= 3

2j

1 +on

- i - i

+ of i

Using the expansion for the exponential,

ex = l + x + 0(x2), x

we get