Why a Geometric Series Leads to a Logarithm
The geometric series is useful not only for representing rational functions. With a suitable substitution and integration, it also produces a power series for \(\ln(1+x)\). This gives a direct way to construct its Maclaurin polynomials and approximate logarithm values.
The interval of convergence matters just as much as the formula. A power series can converge at one endpoint of its interval and diverge at the other, so the endpoint tests must be done separately.
Deriving the Logarithm Series
For \(|t|<1\), the geometric series gives \(\frac{1}{1+t}=1-t+t^2-t^3+\cdots\). The reason this can be used to obtain a logarithm series is that the derivative of \(\ln(1+t)\) is \(\frac{1}{1+t}\), and \(\ln(1+0)=0\). Integrating the geometric series from \(0\) to \(x\) therefore gives the desired function and its coefficients.
To justify the integration and the resulting identity, start with the finite geometric-series formula. For an integer \(N\ge 1\),
Integrating from \(0\) to \(x\) gives
If \(|x|<1\), then along the interval between \(0\) and \(x\), \(1+t\ge 1-|x|>0\). Thus the absolute value of the final integral is at most \(\frac{|x|^{N+1}}{(N+1)(1-|x|)}\), which tends to \(0\) as \(N\) increases. The finite sums therefore approach \(\ln(1+x)\). Replacing \(k+1\) by \(n\) in the index gives the series
This derivation explains the alternating signs and the denominator of each term: integrating \(t^{n-1}\) produces \(x^n/n\).
The Convergence Interval and Its Endpoints
For \(|x|<1\), the series converges absolutely: the absolute values of its terms are \(|x|^n/n\), which are at most \(|x|^n\), and the geometric series \(\sum |x|^n\) converges. The boundary points require separate checks.
At \(x=1\), the series becomes \(1-\frac12+\frac13-\frac14+\cdots\). Its terms decrease in magnitude to \(0\), so the alternating-series test shows that it converges. It is not absolutely convergent, since the absolute-value series is the divergent harmonic series.
The sum at this endpoint is \(\ln 2\), not merely some unspecified limit. In the finite-integral identity above, set \(x=1\). Since \(1+t\ge 1\) on \(0\le t\le1\), the absolute value of the remainder integral is at most \(\int_0^1 t^N\,dt=\frac{1}{N+1}\), which tends to \(0\). Consequently the finite sums converge to \(\ln 2\).
At \(x=-1\), the terms are \(-1-\frac12-\frac13-\frac14-\cdots\), whose partial sums decrease without bound. The series diverges there, and \(\ln(1+x)\) itself is undefined at \(x=-1\) because its argument is zero.
Worked Example: Write the Degree-Five Polynomial
Worked Example: Write the Degree-Five Polynomial
Use the coefficient rule to write \(P_5(x)\), then check its degree and signs.
Start with \(P_m(x)=\sum_{n=1}^{m}\frac{(-1)^{n+1}x^n}{n}\). Set \(m=5\):
Since the signs are \(+,-,+,-,+\), this simplifies to
The last term has degree \(5\), as required. Differentiating also checks the coefficient pattern: \(P_5'(x)=1-x+x^2-x^3+x^4\), the first five terms of the geometric series for \(\frac{1}{1+x}\). This derivative check verifies the signs and denominators in the polynomial.
Approximating Values with Positive Inputs
When \(0<x\le1\), the logarithm series is alternating. Its term magnitudes \(x^n/n\) decrease to \(0\), because \(\frac{x^{n+1}/(n+1)}{x^n/n}=\frac{nx}{n+1}<1\). The alternating-series remainder bound therefore applies: after \(m\) terms, the error is at most the magnitude of the first omitted term, \(x^{m+1}/(m+1)\). The error also has the sign of that first omitted term.
Worked Example: Approximate \(\ln(3/2)\)
Since \(\ln(3/2)=\ln(1+1/2)\), substitute \(x=\frac12\) in the degree-five polynomial:
Using denominator \(960\), the five terms are \(\frac{480}{960}\), \(-\frac{120}{960}\), \(\frac{40}{960}\), \(-\frac{15}{960}\), and \(\frac{6}{960}\). Therefore
The first omitted term is negative, with magnitude \(\frac{(1/2)^6}{6}=\frac{1}{384}\). Thus the error is at most \(\frac{1}{384}\), and the alternating-series sign tells us that the exact value is below \(\frac{391}{960}\). In particular, the symmetric error bound gives
With denominator \(1920\), \(\frac{391}{960}=\frac{782}{1920}\) and \(\frac{1}{384}=\frac{5}{1920}\). The interval is therefore \(\left[\frac{777}{1920},\frac{787}{1920}\right]\), or \(\left[\frac{259}{640},\frac{787}{1920}\right]\). The alternating-series sign gives the sharper upper endpoint \(\frac{391}{960}\); the lower endpoint is the next partial sum, \(\frac{777}{1920}=\frac{259}{640}\). This checks the arithmetic and brackets the logarithm.
Approximating Values with Negative Inputs
For \(-1<x<0\), the series still converges, but all its terms are negative: \(x\), \(-x^2/2\), \(x^3/3\), and so on are each negative. This is not an alternating series, so the alternating-series remainder bound does not apply. A geometric comparison can instead bound the remaining terms.
Worked Example: Approximate \(\ln(3/4)\)
Write \(\ln(3/4)=\ln(1-\frac14)\), so \(x=-\frac14\). The degree-four polynomial gives
Every omitted term is negative. For \(k\ge5\), \(1/k\le1/5\), so the magnitude of the tail is at most
Thus the exact logarithm is below the partial sum and no more than \(\frac{1}{3840}\) below it:
Using denominator \(15360\), this is \(-\frac{4419}{15360}\le\ln(3/4)\le-\frac{4415}{15360}\). The direction of the bounds matches the fact that the omitted terms are negative.
Common Mistake: Ignoring Signs or Endpoints
A frequent error is to substitute \(x=-1\) and claim that the series converges because its terms alternate. They do not alternate there:
All terms are negative, and the harmonic sum diverges. The correction is to test each endpoint in the actual series, rather than infer endpoint behavior from the alternating signs visible for positive \(x\). The left endpoint is excluded; the right endpoint is included.
Another error is to apply the alternating-series remainder bound to \(x<0\). At \(x=-1/4\), for example, the terms do not alternate, so that theorem cannot justify an error estimate. Use a suitable comparison for the tail, as in the geometric bound above.
Why This Series Is Useful
The logarithm series turns an unfamiliar function value into a finite sum of powers and fractions. It is especially convenient when \(x\) is near zero, because the powers \(x^n\) then become small quickly. The convergence interval also gives an immediate restriction: this particular series represents \(\ln(1+x)\) for \(-1<x\le1\), not for every real input.
For approximation work, keep three checks together: confirm that the chosen input is in the convergence interval, calculate the polynomial terms with their signs, and justify the error bound using a test that actually applies to those terms.
Check Your Understanding
Answer these questions using the series and convergence ideas developed above.
- Write the degree-six Maclaurin polynomial for \(\ln(1+x)\).
- Why does the series converge at \(x=1\) but diverge at \(x=-1\)?
- What remainder bound applies after degree \(m\) when \(0<x\le1\)?
- Why is the alternating-series remainder bound not appropriate for \(x=-1/4\)?
- For \(\ln(5/4)\), which value of \(x\) should be substituted into the series?
