📂Calculus

Derivation of the Series Form of the Natural Logarithm and Proof of the Convergence of the Alternating Harmonic Series

Theorem

$$\ln(1-x)=\sum _{ n=0 }^{ \infty }{ \frac { -{ x }^{ n+1 } }{ n+1 } }$$

Description

The series form of $\ln(1-x)$ can be relatively easily obtained. For $\ln(1+x)$, it is enough to substitute $-x$ for $x$ as a result of the theorem.

$$-\ln(1-x)=x+\frac { { x }^{ 2 } }{ 2 }+\frac { { x }^{ 3 } }{ 3 }+\frac { { x }^{ 4 } }{ 4 }+ \cdots$$

If $(-x)$ is substituted for $x$, then

$$-\ln(1+x)=-x+\frac { { x }^{ 2 } }{ 2 }-\frac { { x }^{ 3 } }{ 3 }+\frac { { x }^{ 4 } }{ 4 }- \cdots$$

$$\implies \ln(1+x)=x-\frac { { x }^{ 2 } }{ 2 }+\frac { { x }^{ 3 } }{ 3 }-\frac { { x }^{ 4 } }{ 4 }+ \cdots$$

Meanwhile, by substituting $1$ for $x$ in $\ln(1+x)$, one can understand that an alternating harmonic series converges.

$$\ln 2 = 1- {1 \over 2} + { 1 \over 3} - { 1 \over 4 }+ \cdots$$

Of course, strictly speaking, since it is a series expansion obtained from $-1 < x < 1$, it is necessary to mention that $\ln (1+x)$ is continuous at $x=1$, and then compute the left-hand limit as follows.

$$\begin{align*} \ln 2 =& \ln \lim_{x \to 1} (1 + x) \\ =& \lim_{x \to 1} \ln (1 + x) \\ =& \lim_{x \to 1-} \sum_{n=0}^{\infty} {{(-x)^{n}} \over {n+1}} \end{align*}$$ This fact is primarily used as an example to explain the concept of absolute convergence, where the alternating harmonic series converges as

$$\sum_{n=1}^{\infty} {{(-1)^{n-1}} \over {n}} = 1- {1 \over 2} + { 1 \over 3} - { 1 \over 4 }+ \cdots = \ln 2 < \infty$$

while the series of their absolute values, the harmonic series, diverges, so

$$\sum_{n=1}^{\infty} \left| {{(-1)^{n-1}} \over {n}} \right| = \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n } }=\infty$$

is satisfied. Therefore, it becomes the simplest example to explain that ‘convergence does not necessarily mean absolute convergence.’

Proof

For $-1<x<1$, the sum of the geometric series $\sum_{n=0}^{\infty} x^{n}$ is as follows.

$${{ 1 } \over { 1-x }}=1+x+{ x }^{ 2 }+{ x }^{ 3 }+ \cdots$$

Taking integration on both sides gives

$$-\ln(1-x)=c+x+\frac { { x }^{ 2 } }{ 2 }+\frac { { x }^{ 3 } }{ 3 }+\frac { { x }^{ 4 } }{ 4 }+ \cdots$$

Since when $x=0$, $-\ln(1-0)=0=c+0$, we get $c=0

$$\therefore \ln(1-x)=\sum _{ n=0 }^{ \infty }{ \frac { -{ x }^{ n+1 } }{ n+1 } }$$

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See Also

• Divergence of the harmonic series