Deriving the Logarithm Series From the Geometric Series

From a Geometric Series to a Logarithm

The geometric series represents a rational function using powers of its input. Integrating that representation produces a different pattern: the coefficient of the \(n\)th power becomes \(1/n\). This is the power series for the natural logarithm, and it is useful both for representing logarithmic functions and for checking how power-series operations affect coefficients.

The derivation has an important condition. The geometric series for \(1/(1-t)\) converges when \(|t|<1\). To integrate it term by term, we must also justify that the series behaves well throughout the interval of integration. That justification gives the logarithm identity for every \(x\) strictly between \(-1\) and \(1\).

The Geometric Representation

For \(|t|<1\), the geometric series and its sum satisfy

\frac{1}{1-t}=\sum_{k=0}^{\infty}t^k =1+t+t^2+t^3+\cdots.

Here \(t\) is a real variable. To derive a logarithm series, integrate both sides from \(0\) to \(x\). The left side has antiderivative \(-\ln(1-t)\), since \(1-t>0\) when \(-1<t<1\). The right side can be integrated term by term provided the series converges uniformly on the interval of integration.

Key condition: The geometric representation holds for \(|t|<1\). For a fixed \(x\) with \(|x|<1\), the entire interval between \(0\) and \(x\) lies inside \((-1,1)\). On that interval the geometric series converges uniformly, which is the condition needed for the integration step.

Why Term-by-Term Integration Is Valid

Fix any \(x\) with \(|x|<1\), and choose \(r\) so that \(|x|\leq r<1\). Every \(t\) between \(0\) and \(x\) then satisfies \(|t|\leq r\). The remainder after the terms from \(k=0\) through \(k=N\) is

\frac{1}{1-t}-\sum_{k=0}^{N}t^k =\frac{t^{N+1}}{1-t}.

For \(|t|\leq r\), its absolute value is bounded by

\left|\frac{t^{N+1}}{1-t}\right| \leq \frac{r^{N+1}}{1-r}.

The bound tends to zero as \(N\) tends to infinity and does not depend on \(t\) in the interval. Thus the geometric series converges uniformly there. A uniformly convergent series of continuous functions on a closed, bounded interval may be integrated term by term. This applies to the interval from \(0\) to \(x\), whether \(x\) is positive or negative.

Theorem: For every real \(x\) with \(-1<x<1\),

-\ln(1-x)=\sum_{n=1}^{\infty}\frac{x^n}{n}.

The series is also written with its first few terms as

-\ln(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots, \qquad -1<x<1.

Derivation of the Logarithm Series

Integrate the geometric representation over the interval from \(0\) to \(x\). Uniform convergence, established above, permits integration term by term:

\int_0^x\frac{1}{1-t}\,dt = \int_0^x\sum_{k=0}^{\infty}t^k\,dt = \sum_{k=0}^{\infty}\int_0^x t^k\,dt.

The left side is evaluated using the antiderivative \(-\ln(1-t)\):

\int_0^x\frac{1}{1-t}\,dt = \bigl[-\ln(1-t)\bigr]_0^x = -\ln(1-x)-\bigl(-\ln(1)\bigr) = -\ln(1-x).

For each nonnegative integer \(k\), the integral of the power is

\int_0^x t^k\,dt=\frac{x^{k+1}}{k+1}.

Substituting these integrals into the series gives

-\ln(1-x)=\sum_{k=0}^{\infty}\frac{x^{k+1}}{k+1}.

Now set \(n=k+1\). As \(k\) runs through \(0,1,2,\ldots\), the new index \(n\) runs through \(1,2,3,\ldots\). Therefore

-\ln(1-x)=\sum_{n=1}^{\infty}\frac{x^n}{n}.

The factor \(1/n\) comes directly from integrating the power \(t^{n-1}\). The lower limit \(0\) also makes the constant term vanish: every integrated power contributes a positive power of \(x\).

Worked Examples

Worked Example: Representing \(\ln 2\)

Choose \(x=\frac12\). This is inside the required interval because \(-1<\frac12<1\). The theorem gives

-\ln\left(1-\frac12\right) = \sum_{n=1}^{\infty}\frac{(1/2)^n}{n}.

Since \(1-\frac12=\frac12\) and \(-\ln(\frac12)=\ln 2\), the logarithm has the representation

\ln 2=\sum_{n=1}^{\infty}\frac{1}{n2^n} =\frac12+\frac18+\frac1{24}+\frac1{64}+\frac1{160}+\cdots.

For example, the sum of the first five terms is found with denominator \(960\):

\frac12+\frac18+\frac1{24}+\frac1{64}+\frac1{160} = \frac{480+120+40+15+6}{960} = \frac{661}{960}.

As a check, substituting \(x=\frac12\) into the general identity gives \(-\ln(1-\frac12)=\ln 2\), and the first five terms computed from \(x^n/n\) are exactly those displayed. The finite sum is a partial sum of the series, not an equality to the full logarithm.

Worked Example: Representing \(\ln(3/2)\)

Choose \(x=-\frac12\), which also satisfies \(-1<x<1\). Substitution gives

-\ln\left(1-\left(-\frac12\right)\right) = \sum_{n=1}^{\infty}\frac{(-1/2)^n}{n}.

The logarithm on the left is \(-\ln(\frac32)\). Multiplying the identity by \(-1\) therefore yields

\ln\left(\frac32\right) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n2^n} = \frac12-\frac18+\frac1{24}-\frac1{64}+\cdots.

The first four terms add to

\frac12-\frac18+\frac1{24}-\frac1{64} = \frac{96-24+8-3}{192} = \frac{77}{192}.

As a check, \(1-(-\frac12)=\frac32\), so the substituted left side before multiplying by \(-1\) is \(-\ln(\frac32)\); the resulting identity has the positive logarithm and alternating signs shown. The arithmetic for the four-term partial sum uses \(1/2=96/192\), \(1/8=24/192\), \(1/24=8/192\), and \(1/64=3/192\).

Worked Example: Representing \(\ln 3\)

Set \(x=\frac23\). Since \(0<\frac23<1\), the logarithm series applies:

-\ln\left(1-\frac23\right) = \sum_{n=1}^{\infty}\frac{(2/3)^n}{n}.

The left side is \(-\ln(\frac13)=\ln 3\). The first four terms on the right are

\frac23+\frac{(2/3)^2}{2}+\frac{(2/3)^3}{3}+\frac{(2/3)^4}{4} = \frac23+\frac29+\frac8{81}+\frac4{81}.

Writing each term with denominator \(81\) gives

\frac{54}{81}+\frac{18}{81}+\frac8{81}+\frac4{81} = \frac{84}{81} = \frac{28}{27}.

Thus \(\ln 3\) is represented by the infinite series \(\sum_{n=1}^{\infty}(2/3)^n/n\), whose first four terms sum to \(28/27\). As a check, \(1-\frac23=\frac13\), and so the left side of the theorem is \(-\ln(\frac13)=\ln 3\). The four terms also follow directly by substituting \(x=\frac23\) into \(x^n/n\).

Checking the Coefficients by Differentiation

A useful check is to differentiate the series. This is valid at interior points: on every interval \([-r,r]\) with \(r<1\), the differentiated terms are \(x^{n-1}\), and their absolute values are bounded by \(r^{n-1}\). The geometric majorant \(\sum_{n=1}^{\infty}r^{n-1}\) converges, so the derivative series converges uniformly on that interval. The power-series differentiation theorem therefore permits term-by-term differentiation for \(|x|<1\).

Differentiate the proposed logarithm series:

\frac{d}{dx}\left(\sum_{n=1}^{\infty}\frac{x^n}{n}\right) = \sum_{n=1}^{\infty}\frac{d}{dx}\left(\frac{x^n}{n}\right) = \sum_{n=1}^{\infty}x^{n-1} = 1+x+x^2+x^3+\cdots = \frac{1}{1-x}.

This agrees with the derivative of the function on the left:

\frac{d}{dx}\bigl[-\ln(1-x)\bigr] = -\frac{1}{1-x}(-1) = \frac{1}{1-x}.

The derivatives agree for every \(|x|<1\). The two functions also have the same value at \(x=0\): both are zero. Since two differentiable functions on an interval with equal derivatives and equal values at one point are equal throughout that interval, this independently verifies the identity and confirms the coefficients \(1/n\).

Start with the geometric series.
For \(|x|<1\), \(1/(1-x)=\sum_{k=0}^{\infty}x^k\).

Integrate from zero.
Uniform convergence on the interval between \(0\) and \(x\) justifies integrating each term.

Reindex.
Each integral \(x^{k+1}/(k+1)\) becomes \(x^n/n\) when \(n=k+1\).

Differentiate to check.
The derivative series is the original geometric series, and both sides vanish at zero.

Common Pitfall: Losing a Power or a Sign

A frequent error is to integrate \(t^k\) as \(t^k/(k+1)\), forgetting that the exponent increases by one. That would incorrectly produce

\sum_{k=0}^{\infty}\frac{x^k}{k+1}.

The correct integration is

\int_0^x t^k\,dt = \left[\frac{t^{k+1}}{k+1}\right]_0^x = \frac{x^{k+1}}{k+1}.

The extra power matters: the series for \(-\ln(1-x)\) has no constant term and begins with \(x\). A separate sign error can occur when evaluating the integral. Since the derivative of \(-\ln(1-t)\) is \(1/(1-t)\), the integral from \(0\) to \(x\) is \(-\ln(1-x)\), not \(\ln(1-x)\).

Scope of the Identity

The established identity holds for \(-1<x<1\). This restriction is part of the result: the geometric representation and the uniform-convergence argument used in the derivation apply to interior intervals contained in \((-1,1)\). The argument does not by itself determine what happens at the endpoints. In particular, the endpoint \(x=1\) is not in the domain of \(-\ln(1-x)\); endpoint behavior should not be inferred by simply substituting \(x=1\) into the interior identity.

Within its interval, the series is a reliable representation: its coefficients can be checked by differentiation, and substitutions give series for logarithms such as \(\ln 2\), \(\ln(3/2)\), and \(\ln 3\). The central habits are to track the integration limits and indices carefully and to retain the convergence condition throughout the calculation.

Check Your Understanding

Use the derivation and its convergence conditions to answer the following.

Why does integrating the term \(t^k\) from \(0\) to \(x\) produce \(x^{k+1}/(k+1)\), rather than \(x^k/(k+1)\)?
What uniform-convergence fact justifies integrating the geometric series term by term for a fixed \(x\) with \(|x|<1\)?
Substitute \(x=\frac13\) into the logarithm series and write the first four terms representing \(-\ln(2/3)\).
Differentiate \(\sum_{n=1}^{\infty}x^n/n\) for \(|x|<1\). Which geometric series results?
Why does the derivation not justify substituting \(x=1\) into the identity?

Deriving the Logarithm Series From the Geometric Series

What You'll Learn

From a Geometric Series to a Logarithm

The Geometric Representation

Why Term-by-Term Integration Is Valid

Derivation of the Logarithm Series