The Logarithm Series

From the Exponential Series to the Logarithm

The Exponential Series tutorial defined the function \(E\) by its power series and established that \(E'(x)=E(x)\), that \(E(x)>0\), and that \(E(x+y)=E(x)E(y)\). These facts allow us to define the natural logarithm as the inverse of \(E\). We will then derive its power series from a finite geometric identity. The resulting expansion is centered at \(0\) when written as a power series in \(x\) for \(\ln(1+x)\) (equivalently, centered at \(y=1\) for \(\ln y\)), and its two boundary points behave differently.

First, \(E\) is strictly increasing: its derivative \(E'(x)=E(x)\) is positive everywhere, so the Mean Value Theorem implies \(E(u)<E(v)\) whenever \(u<v\). Also, the exponential series gives \(E(1)>1\), since its first two terms are \(1+1\) and all its remaining terms are positive. The multiplication law implies \(E(n)=E(1)^n\) for positive integers \(n\), so \(E(n)\) tends to infinity. Since \(E(n)E(-n)=E(0)=1\), we have \(E(-n)=1/E(n)\), which tends to zero. Continuity and the Intermediate Value Theorem now show that the range of \(E\) is exactly \((0,\infty)\).

We need the derivative of this inverse to connect the logarithm to a geometric series. The inverse is continuous: if \(E(a)=y\), then for any \(\varepsilon>0\), \(E(a-\varepsilon)<y<E(a+\varepsilon)\). Values sufficiently close to \(y\) remain between these two bounds, and their inverse images therefore lie between \(a-\varepsilon\) and \(a+\varepsilon\). For \(h\) near zero with \(y+h>0\), put \(u=\ln(y+h)-\ln y\). Continuity gives \(u\to0\) as \(h\to0\), and the definition of the inverse gives \(h=E(a+u)-E(a)\). Since \(E'(a)=E(a)=y\), the difference quotient yields

Consequently, \((\ln y)'=1/y\) for \(y>0\). The Fundamental Theorem of Calculus, applied on the interval between \(1\) and \(1+x\), gives

Deriving the Logarithm Series

For \(t\neq-1\), a finite geometric identity gives, for every integer \(N\geq0\),

Integrate this identity from \(0\) to \(x\). The integral of the finite sum is explicit, and the remainder is an ordinary integral, so no interchange of an infinite sum and an integral is needed. We obtain

This identity both suggests the series and gives a direct way to verify its sum. If \(|x|<1\), then along the interval joining \(0\) and \(x\), \(1+t\geq1-|x|>0\). Therefore the absolute value of the remainder is at most

For fixed \(|x|<1\), this bound tends to zero as \(N\) tends to infinity. At \(x=1\), the denominator \(1+t\) is at least \(1\) on the integration interval, so the remainder instead satisfies

which also tends to zero. We have proved the following result.

Proof. The finite identity and remainder estimates above show that the partial sums tend to \(\ln(1+x)\) when \(|x|<1\) and when \(x=1\). For \(|x|<1\), absolute convergence follows from comparison with the geometric series, since \[ \sum_{n=1}^{\infty}\left|(-1)^{n+1}\frac{x^n}{n}\right| \leq\sum_{n=1}^{\infty}|x|^n<\infty. \] At \(x=1\), the series is the alternating harmonic series \(\sum_{n=1}^{\infty}(-1)^{n+1}/n\). Its terms decrease to zero, so the Alternating Series Test gives convergence. It is not absolutely convergent because \(\sum 1/n\) diverges, by the Cauchy Condensation Test. At \(x=-1\), each term is \(-1/n\), so the partial sums diverge. Finally, the coefficients have absolute values \(1/n\), and \((1/n)^{1/n}\to1\). The Cauchy–Hadamard Formula therefore gives radius \(1\). \(\square\)

Using the Series to Estimate Logarithms

For \(0<x\leq1\), the terms \(x^n/n\) decrease to zero: the ratio of consecutive terms is \(xn/(n+1)\), which is less than \(1\). The Alternating Series Test and its remainder bound therefore provide simple enclosures. For negative \(x\) with \(|x|<1\), all the terms in the logarithm series have the same sign. A geometric comparison can instead bound the tail. These approaches are useful in different parts of the interval.

What the Boundary Behavior Tells Us

The series has radius \(1\), but the radius alone does not decide what happens at either endpoint. At \(x=1\), alternating signs yield convergence, and the remainder can be bounded by the first omitted term. At \(x=-1\), every term is negative and the series becomes the divergent harmonic series with a minus sign. This is why the real logarithm identity above holds on \((-1,1]\), not at \(x=-1\): there \(1+x=0\), outside the domain of \(\ln\).

A common pitfall is to assume that convergence at one endpoint implies convergence at the other, or that a power series converges at every point within its radius and both endpoints as well. The Cauchy–Hadamard Formula determines the radius, while endpoint analysis must be done separately. Another useful distinction is between representing a function and estimating it: the finite geometric identity proves the representation, whereas the alternating-series or geometric-tail bounds provide quantitative error estimates.