log (1 + x) = ∑ (n=1,∞) [(-1)n+1xn]/n

In today's blog, I use the fundamental theorem of calculus and the basic properties of derivatives to show that:
log (1 + x) = ∑ (n=1,∞) [(-1)ⁿ⁺¹xⁿ]/n

First, we need a lemma:

Lemma 1: 1/(x+1) = ∑ (i=1,∞) (-1)⁽ⁱ⁺¹⁾*x^i-1

for x ≠ -1

Proof:

(1) Let s = ∑ (i=1,∞) (-1)⁽ⁱ⁺¹⁾*x^i-1 which gives us:

s = 1 - x + x² -x³ + ....

(2) xs = x - x² + x³ - x⁴ + ...

(3) Then xs + s = 1

(4) Solving for s gives us:

s(x+1) = 1

so that:

s = 1/(x+1)

QED

Theorem: log (1 + x) = ∑ (n=1,∞) [(-1)ⁿ⁺¹xⁿ]/n

Proof:

(1) From Lemma 1 above;

1/(1+x) = ∑ (i=1,∞) (-1)⁽ⁱ⁺¹⁾*x^i-1

(2) d/dx(ln(1+x)) = 1/(1+x) since:

(a) Let g(x) = 1 + x

(b) Let h(x) = ln(x)

(c) Using the Chain Rule for Derivatives (see Lemma 2, here), we have:

d/dx(ln(1+x)) = d/dx(h(g(x)) = h'(g(x))*g'(x)

where:

h'(g(x)) = ln(1+x) = 1/(1+x) [see Lemma 1, here]

g'(x) = d/dx(1 + x) = d/dx(1) + d/dx(x) = 0 + 1 = 1

so:

d/dx(ln(1+x)) = [1/(1+x)]*1 = 1/(1+x)

(3) Since d/dx(ln(1+x)) = 1/(1+x), we know that ∫(1/(1+x)dx) = ln(1+x)+C [See Theorem 2, here]

(4) If we take the antiderivative of both sides (see Definition 2, here for definition of antiderivative), we have:

ln(1+x) = ∫(1 - x + x² - x³ + ...)

(5) Using the linearity property of integrals (see Lemma, here) gives us:

ln(1+x) = ∫1 - ∫xdx + ∫x²dx - ∫x³dx + ...

(6) Using the power rule for integrals (see Lemma, here) gives us:

ln(1+x) = x - x²/2 + x³/3 -x⁴/4 + ....

QED