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
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