Advanced High-School Mathematics

302 CHAPTER 5 Series and Differential Equations

In this case, as long as −^12 ≤ x ≤ 1, then we are guaranteed that
∣∣
∣∣
∣

xn+1

(1 +c)n+1

∣∣

∣∣

∣ ≤ 1. Therefore, as n → ∞ we have

∣∣

∣∣

∣

xn+1

(1+c)n+1(n+1)

∣∣

∣∣

∣ → 0.

Therefore, we at least know that for if−^12 ≤x≤1,

ln(1 +x) =x−

x^2

2

+

x^3

3

−···=

∑∞

n=1

(−1)n−^1

xn

n

.

In particular, this proves the fact anticipated on page 278, viz., that

ln(2) = 1−

1

2

+

1

3

−···=

∑∞

n=1

(−1)n−^1

1

n

.

Example 3. One easily computes that the first two terms of the

Maclaurin series expansion of

√

1 +x is 1 +

x

2

. Let’s give an upper

bound on the error
∣∣
∣∣
∣

√

1 +x−

Ç

1 +

x

2

å∣∣

∣∣

∣

when|x|< 0. 01 .ByTaylor’s Theorem with Remainder, we know

that the absolute value of the error is given by

∣∣

∣∣

∣f

′′(c)x

2

2

∣∣

∣∣

∣, where c is

between 0 andx, and wheref(x) =

√

1 +x. Sincef′′(c) =

− 1

4(1 +c)^3 /^2

,

and sincecis between 0 andx, we see that 1 +c≥.99 and so

∣∣

∣∣

∣f

′′(c)

∣∣

∣∣

∣=

1

4(1 +c)^3 /^2

≤

1

4 ×. 993 /^2

≤. 254.

This means that the error in the above approximation is no more than

∣∣

∣∣

∣f

′′(c)x

2

2

∣∣

∣∣

∣≤.^254 ×

(0.01)^2

2

<. 000013.

Another way of viewing this result is that,accurate to four decimal
places,

√

1 +x= 1 +

x

2

whenever|x|< 0. 01.

Exercises

1. Show that for allx, ex=

∑∞

n=0

xn

n!

.