From Standard Series to New Representations
A few familiar functions have Maclaurin series that can be derived from the geometric series. These expansions are useful not only for approximating values: they also represent functions exactly on specified intervals. The interval matters. A power series can converge at one endpoint, diverge at the other, and still represent the function at every point between them.
Here we derive the series for \(\ln(1+x)\) and \(\arctan x\), then state and use the generalized binomial series for fractional powers of \(1+x\). For each series, we will check the endpoints separately rather than assuming that convergence at interior points settles the endpoint question.
The Logarithmic Series
For \(t\ne -1\), a finite geometric-series identity gives
When \(|t|<1\), the last term tends to zero as \(N\) increases, giving the geometric expansion of \(1/(1+t)\). Integrating the finite identity from \(0\) to \(x\), then taking the limit for \(|x|<1\), gives the logarithmic series. The integral of the remainder tends to zero: for fixed \(|x|<1\), its absolute value is at most \(\frac{|x|^{N+2}}{(N+2)(1-|x|)}\).
At \(x=1\), the series is \(1-\frac12+\frac13-\frac14+\cdots\). Its terms decrease in magnitude to zero, so the alternating-series test gives convergence. The sum is \(\ln 2\): for example, the integrated finite geometric identity above also applies at \(x=1\), since the absolute value of its integrated remainder is at most \(\int_0^1 t^{N+1}\,dt=\frac{1}{N+2}\), which tends to zero.
At \(x=-1\), the function \(\ln(1+x)\) is undefined. The series terms are \(\frac{(-1)^n(-1)^{n+1}}{n+1}=-\frac{1}{n+1}\), so the series becomes the negative harmonic series and diverges, even though its terms tend to zero. The term test alone does not show divergence here; the divergence follows from the harmonic series.
Worked Example: Expanding and Checking \(\ln(3/2)\)
Substitute \(x=\frac12\) into the logarithmic series. Since \(-1<\frac12<1\), the series represents the function:
Compute the first four terms one at a time: \(\frac12\); \(-\frac{1}{4\cdot2}=-\frac18\); \(\frac{1}{8\cdot3}=\frac1{24}\); and \(-\frac{1}{16\cdot4}=-\frac1{64}\). Their sum is
Thus the first four terms give the partial sum \(\frac{77}{192}\), while the full series equals \(\ln(3/2)\). As a check on the expansion itself, its general term is \(\frac{(-1)^n x^{n+1}}{n+1}\), exactly the term obtained by integrating \((-1)^n x^n\) from the geometric series. The substitution is inside the interval of validity.
The Arctangent Series
Replace \(t\) by \(t^2\) in the geometric expansion of \(1/(1+t)\), then integrate from \(0\) to \(x\). Since the derivative of \(\arctan t\) is \(1/(1+t^2)\) and \(\arctan 0=0\), the result is
The endpoints can also be checked directly using the finite geometric identity. After integrating its remainder with \(t^2\) in place of \(t\), the remainder at \(x=1\) has absolute value at most \(\int_0^1 t^{2N+2}\,dt=\frac{1}{2N+3}\), which tends to zero. At \(x=-1\), the same bound applies after reversing the integration limits. Therefore the series converges to \(\arctan x\) at both endpoints. Equivalently, the endpoint series are alternating series with terms tending to zero.
Worked Example: Evaluating the Arctangent Series at an Endpoint
Set \(x=1\). The \(n\)th term is \(\frac{(-1)^n}{2n+1}\), so the series is
Its term magnitudes are \(\frac1{2n+1}\): they decrease to zero, so the alternating-series test confirms convergence. The endpoint remainder estimate from the integrated geometric identity is at most \(\frac{1}{2N+3}\) after the terms through \(n=N\), and this bound tends to zero. Since the series represents \(\arctan x\) at \(x=1\), its sum is \(\arctan(1)=\frac{\pi}{4}\). Replacing \(x\) by \(-1\) changes every odd power’s sign, so the series sums to \(-\frac{\pi}{4}=\arctan(-1)\) there as well.
The Generalized Binomial Series
For a real exponent \(\alpha\), define the generalized binomial coefficient by \(\binom{\alpha}{0}=1\) and, for \(n\ge1\),
If \(\alpha\) is not a nonnegative integer, the generalized binomial series is
The coefficients come from the derivatives of \((1+x)^\alpha\) at \(x=0\). The ratio of the magnitudes of consecutive terms, when defined, tends to \(|x|\); the ratio test therefore gives radius of convergence \(1\). The generalized binomial theorem states that the resulting series equals \((1+x)^\alpha\) throughout \(|x|<1\). For a nonnegative integer exponent, the series instead terminates and is an ordinary polynomial.
The endpoints are not covered by the ratio test, so they must be checked separately. For noninteger \(\alpha\), the coefficient magnitudes have the large-\(n\) behavior of a constant times \(n^{-\alpha-1}\). This gives the following endpoint classification:
| Exponent | At \(x=1\) | At \(x=-1\) | Interval of validity |
|---|---|---|---|
| \(\alpha>0\) | Converges absolutely | Converges absolutely | \([-1,1]\) |
| \(-1<\alpha<0\) | Converges conditionally | Diverges | \((-1,1]\) |
| \(\alpha<-1\) | Diverges | Diverges | \((-1,1)\) |
For \(-1<\alpha<0\), the coefficients at \(x=1\) alternate in sign and their magnitudes decrease to zero, giving conditional convergence. At \(x=-1\), multiplication by \((-1)^n\) removes that alternation; the terms have one sign and magnitudes comparable to \(n^{-\alpha-1}\), whose exponent lies between \(0\) and \(1\), so the series diverges. When \(\alpha<-1\), the term magnitudes at either endpoint do not tend to zero. For \(\alpha>0\), the endpoint series converge absolutely; at \(x=-1\) their sum is \(0=(1-1)^\alpha\).
Worked Example: Expanding the Square Root
Take \(\alpha=\frac12\). The coefficient formula gives \(c_0=1\), \(c_1=\frac12\), and then
Thus the expansion begins
Here \(\alpha>0\), so both endpoints are included. At \(x=1\), the first five terms sum to
This is a partial sum; the full convergent series equals \(\sqrt2\). At \(x=-1\), the series also converges absolutely and its sum is \(\sqrt{1-1}=0\). These endpoint checks agree with the interval \([-1,1]\) for this positive fractional exponent.
Why the Endpoints Matter
A power series’ radius tells us what happens at points strictly inside or outside the radius, but it does not determine what happens at the boundary. The logarithmic series illustrates the difference sharply: it converges at \(x=1\), yet diverges at \(x=-1\). The arctangent series converges at both endpoints. For a fractional power, endpoint behavior depends on the exponent.
A common mistake is to say that the logarithmic series at \(x=-1\) diverges because its terms fail to approach zero. That calculation is incorrect: each term is \(-1/(n+1)\), which does approach zero. The correct reason is that the partial sums are the negatives of harmonic partial sums, and those grow without bound in magnitude. Always calculate the endpoint term before applying a convergence test.
Check Your Understanding
Use the formulas and endpoint reasoning above to answer each question.
- What is the interval of convergence of the series for \(\ln(1+x)\), and what series results at \(x=-1\)?
- Why does the arctangent series converge at \(x=1\), and what is its sum there?
- For \(\alpha=\frac12\), are both endpoints included in the generalized binomial series? Explain using the endpoint classification.
- For \(-1<\alpha<0\), which binomial-series endpoint converges, and is that convergence absolute or conditional?
- Why is it not enough to use the ratio test to decide whether a power series converges at an endpoint?
