The Logarithm Series From Termwise Integration

From a Geometric Series to a Logarithm

The geometric series is useful not only for representing rational functions. Integrating it term by term can produce power series for other functions. In this tutorial, the function is the natural logarithm.

For \(|x|<1\), the geometric series gives

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

The derivative of \(\ln(1+x)\) is \(1/(1+x)\), and \(\ln(1)=0\). These facts suggest integrating from \(0\) to \(x\). The key point is that the geometric series may be integrated term by term on any interval that stays strictly inside \((-1,1)\). This gives the series when \(|x|<1\); the endpoints must then be checked separately.

Definition: The logarithm series is the power series \(\displaystyle \sum_{n=0}^{\infty}(-1)^n\frac{x^{n+1}}{n+1}\). For \(|x|<1\), it represents \(\ln(1+x)\).

Why Termwise Integration Works Inside the Interval

Fix \(x\) with \(|x|<1\). For every \(t\) between \(0\) and \(x\), choose \(r\) such that \(|x|\le r<1\). Then \(|t|\le r\), and the remainder after \(N+1\) terms of the geometric series satisfies

\left| \frac{1}{1+t}-\sum_{n=0}^{N}(-1)^n t^n \right| = \left|\frac{(-t)^{N+1}}{1+t}\right| \le \frac{r^{N+1}}{1-r}.

The bound tends to zero as \(N\) increases and does not depend on \(t\) in the interval between \(0\) and \(x\). Thus the geometric partial sums converge uniformly there. Uniform convergence on this finite interval permits integration term by term. This argument applies whether \(x\) is positive or negative, as long as \(|x|<1\).

Start with the geometric series.
For \(|t|<1\), write \(1/(1+t)=\sum_{n=0}^{\infty}(-1)^n t^n\).

Integrate on an interior interval.
For fixed \(|x|<1\), uniform convergence between \(0\) and \(x\) justifies integrating the series term by term.

Use the logarithm’s value at zero.
The integral of \(1/(1+t)\) from \(0\) to \(x\) is \(\ln(1+x)\), since \(x>-1\).

Carrying out those integrations gives the identity

\begin{aligned} \ln(1+x) &=\int_0^x \frac{1}{1+t}\,dt\\ &=\int_0^x \sum_{n=0}^{\infty}(-1)^n t^n\,dt\\ &=\sum_{n=0}^{\infty}(-1)^n\int_0^x t^n\,dt\\ &=\sum_{n=0}^{\infty}(-1)^n\frac{x^{n+1}}{n+1}, \qquad |x|<1. \end{aligned}

The integral formula uses \(x>-1\), so that \(1+t\) stays positive along the interval of integration. The series conclusion above is established for \(|x|<1\). It does not yet say what happens at \(x=1\) or \(x=-1\).

The Logarithm Series

Reindexing the terms by setting \(k=n+1\) gives a familiar form.

Theorem: For \(-1<x<1\), \(\displaystyle \ln(1+x)=\sum_{k=1}^{\infty}(-1)^{k+1}\frac{x^k}{k} =x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots.\)

The signs alternate, and the denominator of the \(k\)th term is \(k\). The first term is \(x\), not \(1\), because integrating \(t^n\) produces \(x^{n+1}/(n+1)\). The series is centered at \(x=0\), where \(\ln(1+x)=0\).

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

Set \(x=1/2\), which lies inside the interval of validity. The series becomes

\ln\left(\frac32\right) =\frac12-\frac{(1/2)^2}{2}+\frac{(1/2)^3}{3} -\frac{(1/2)^4}{4}+\cdots.

The first four terms are

\begin{aligned} S_3 &=\frac12-\frac18+\frac1{24}-\frac1{64}\\ &=\frac{96}{192}-\frac{24}{192} +\frac{8}{192}-\frac{3}{192}\\ &=\frac{77}{192} \approx 0.4010417. \end{aligned}

The terms decrease in magnitude and alternate in sign. The alternating-series remainder bound says that the error after these four terms is at most the magnitude of the next term:

\left|\ln\left(\frac32\right)-S_3\right| \le \frac{(1/2)^5}{5} =\frac1{160} =0.00625.

Thus this partial sum gives an approximation with error no greater than \(0.00625\). As a check on the expression being approximated, differentiating the series term by term for \(|x|<1\) recovers \(1/(1+x)\), and the series is zero at \(x=0\); these are the derivative and initial value of \(\ln(1+x)\).

Convergence Inside and Outside the Interval

For \(|x|<1\), the series converges absolutely: the magnitudes of its terms are \(|x|^k/k\), which are bounded above by the terms \(|x|^k\) of a convergent geometric series. At the endpoints, that argument no longer applies. Each endpoint must be tested directly.

If \(|x|>1\), the terms do not approach zero. Indeed,

\left|(-1)^{k+1}\frac{x^k}{k}\right| =\frac{|x|^k}{k} \not\longrightarrow 0 \qquad (|x|>1).

A necessary condition for a series to converge is that its terms tend to zero, so the logarithm series diverges for \(|x|>1\). This leaves the two boundary points \(x=1\) and \(x=-1\).

The Endpoint \(x=1\)

Substituting \(x=1\) gives the alternating harmonic series

1-\frac12+\frac13-\frac14+\frac15-\cdots.

Its terms decrease in magnitude to zero, so the alternating-series test shows that the series converges. It does not converge absolutely, since the absolute-value series is the divergent harmonic series \(\sum_{k=1}^{\infty}1/k\). To confirm that its sum is \(\ln 2\), use the finite geometric identity inside the integral.

For \(0\le t\le1\),

\frac1{1+t} =\sum_{n=0}^{N}(-t)^n+\frac{(-t)^{N+1}}{1+t}.

Integrate from \(0\) to \(1\). Since the absolute value of the integrated remainder is at most \(\int_0^1 t^{N+1}\,dt=1/(N+2)\), it tends to zero. Therefore,

\begin{aligned} \ln 2 &=\int_0^1\frac1{1+t}\,dt\\ &=\lim_{N\to\infty}\sum_{n=0}^{N}\frac{(-1)^n}{n+1}\\ &=1-\frac12+\frac13-\frac14+\cdots. \end{aligned}

Worked Example: Checking the Series at \(x=1\)

The partial sum through the term \(-1/4\) is

1-\frac12+\frac13-\frac14 =\frac{12-6+4-3}{12} =\frac7{12}.

The next term is \(1/5\). The alternating-series remainder bound gives

0<\ln 2-\frac7{12}<\frac15.

The error is positive because the first omitted term is positive. The endpoint series does converge to \(\ln 2\), as shown by the integrated finite geometric identity above; the bound also confirms that this partial sum lies below its sum and within \(1/5\) of it.

The Endpoint \(x=-1\)

At \(x=-1\), every term is negative:

(-1)^{k+1}\frac{(-1)^k}{k}=-\frac1k.

Consequently the series becomes \(-\sum_{k=1}^{\infty}1/k\), which diverges. Its partial sums decrease without bound. Also, \(\ln(1+x)\) is not defined as a real number at \(x=-1\), because its argument is zero. This endpoint is not included in the interval of convergence.

Worked Example: Why \(x=-1\) Is Excluded

The first four terms obtained by substituting \(x=-1\) are

-1-\frac12-\frac13-\frac14 =-\left(\frac{12+6+4+3}{12}\right) =-\frac{25}{12}.

After \(N\) terms, the partial sum is \(-\sum_{k=1}^{N}1/k\). The harmonic sums grow without bound, so these partial sums do not approach a finite value. This verifies divergence at \(x=-1\); it cannot be treated as a value of the real logarithm series.

The Interval of Convergence and a Common Pitfall

Combining the interior result, the test for \(|x|>1\), and the two endpoint checks gives the complete interval:

Takeaway: The series \(\displaystyle \sum_{k=1}^{\infty}(-1)^{k+1}\frac{x^k}{k}\) converges for \(-1<x\le1\). It equals \(\ln(1+x)\) throughout that interval. Convergence is absolute for \(-1<x<1\), conditional at \(x=1\), and absent at \(x=-1\).

A common mistake is to assume that because a power series converges for \(|x|<1\), both endpoints must also be included. For example, writing “the interval is \([-1,1]\)” based only on the radius is incorrect: at \(x=-1\), the series is \(-1-1/2-1/3-\cdots\) and diverges. Conversely, excluding both endpoints without checking would miss the convergence at \(x=1\). The radius identifies where to investigate; endpoint tests determine whether either boundary point belongs.

The endpoint behavior is also a useful reminder that a formula obtained by termwise integration on interior intervals cannot automatically be extended to the boundary. Uniform convergence was justified on intervals strictly inside \((-1,1)\), not on an interval reaching \(x=-1\). At \(x=1\), convergence and the value require their own argument.

Check Your Understanding

Use the derivation and endpoint tests to answer the following questions.

Why can the geometric series be integrated term by term for a fixed \(x\) with \(|x|<1\)?
What is the first four-term partial sum for \(\ln(1+1/3)\)?
Why does the logarithm series converge at \(x=1\) but not absolutely?
What series results from substituting \(x=-1\), and why does it diverge?
Why does the radius of convergence alone not determine whether \(x=1\) or \(x=-1\) is included?