From Taylor Coefficients to a General Binomial Series
For a nonnegative integer exponent, the binomial theorem gives a finite polynomial for \((1+x)^\alpha\). For a real exponent that is not a nonnegative integer, the Taylor expansion has infinitely many terms. Its coefficients still follow a systematic pattern, but convergence at the two endpoints requires separate attention.
The goal is to derive those coefficients, find where the resulting series represents the function, and distinguish the endpoint behavior for different values of \(\alpha\). The exceptional case \(\alpha=0\) matters: its series terminates, so it must not be classified like the nonterminating series.
Deriving the Coefficients
Let \(f(x)=(1+x)^\alpha\), defined for \(x>-1\). Repeated differentiation gives, for \(n\geq 1\),
At \(x=0\), this gives \(f^{(n)}(0)/n!=\binom{\alpha}{n}\). Since the Taylor coefficient of degree \(n\) is \(f^{(n)}(0)/n!\), the Taylor series about zero must have these generalized binomial coefficients.
The symbol \(\sim\) here indicates the Taylor series whose convergence and equality to the function will be established below. The coefficients satisfy a useful recurrence obtained by dividing consecutive products:
If \(\alpha=m\) is a nonnegative integer, the factor \(\alpha-n\) becomes zero at \(n=m\). All coefficients after degree \(m\) are zero, and the series is the finite binomial polynomial. If \(\alpha\) is not a nonnegative integer, none of the coefficients vanish and the series is nonterminating.
Worked Example: The Exponent Three Halves
Take \(\alpha=\frac32\). Starting with \(\binom{3/2}{0}=1\), use the recurrence to compute the first coefficients:
Thus the series begins
The exponent is not a nonnegative integer, so the series does not terminate. The radius calculation below gives radius \(1\). Since \(\alpha>0\), the series converges absolutely at both endpoints. Its endpoint sums are \(2^{3/2}=2\sqrt2\) at \(x=1\) and \(0\) at \(x=-1\), agreeing with the corresponding function values. This checks both the coefficient pattern and the endpoint values.
Radius of Convergence
For a nonterminating series, the ratio of consecutive coefficient magnitudes is
For the terms \(\binom{\alpha}{n}x^n\), the ratio test therefore gives a limiting ratio \(|x|\). The series converges absolutely when \(|x|<1\) and diverges when \(|x|>1\). Its radius of convergence is \(1\). At \(|x|=1\), the ratio test does not decide convergence; those two endpoints require a separate analysis.
Why the Series Represents the Function Inside Its Radius
For completeness, let \(S(x)=\sum_{n=0}^{\infty}\binom{\alpha}{n}x^n\) for \(|x|<1\), in the nonterminating case. A power series may be differentiated term by term strictly inside its radius of convergence. The coefficient recurrence then implies a differential equation for \(S\).
The coefficient of \(x^n\) in \(S'(x)\) is \((n+1)\binom{\alpha}{n+1}\). By the recurrence, this is \((\alpha-n)\binom{\alpha}{n}\).
The coefficient of \(x^n\) in \(xS'(x)\) is \(n\binom{\alpha}{n}\). Thus the coefficient of \(x^n\) in \((1+x)S'(x)\) is \(\alpha\binom{\alpha}{n}\), and \((1+x)S'(x)=\alpha S(x)\).
On \((-1,1)\), the function \(g(x)=(1+x)^\alpha\) satisfies \((1+x)g'(x)=\alpha g(x)\) and \(g(0)=1\). The quotient \(S(x)/g(x)\) has derivative zero there, and its value at zero is \(1\). Hence \(S(x)=g(x)\) throughout \((-1,1)\).
This verifies that the convergent series is not merely a list of Taylor coefficients: throughout its open interval of convergence it equals the intended function.
A Coefficient Estimate for the Endpoints
To analyze the endpoints, it is useful to know how large the coefficients are. Suppose \(\alpha\) is not a nonnegative integer, so the coefficients do not terminate. For all sufficiently large \(n\), the recurrence gives
Taking logarithms of the product of these ratios shows that \(\log|\binom{\alpha}{n}|=-(\alpha+1)\log n+O(1)\). Indeed, for large \(n\), the logarithm of each ratio is \(-(\alpha+1)/n+O(1/n^2)\); the sum of the \(1/n^2\) terms stays bounded, while \(\sum_{k=1}^n 1/k=\log n+O(1)\). Consequently there are positive constants \(C_1,C_2\), depending on \(\alpha\), such that for all sufficiently large \(n\),
This comparison with a power of \(n\), together with the coefficient signs, determines the endpoint behavior.
What Happens at \(x=1\) and \(x=-1\)?
For positive \(\alpha\), the coefficient magnitudes are comparable to \(n^{-\alpha-1}\), whose sum converges because \(\alpha+1>1\). Thus both endpoint series converge absolutely. This also holds for positive integer \(\alpha\), when the series terminates.
For negative \(\alpha\), every factor in the numerator defining \(\binom{\alpha}{n}\) is negative. Therefore \(\binom{\alpha}{n}\) has sign \((-1)^n\). At \(x=1\), the series alternates. If \(-1<\alpha<0\), its term magnitudes decrease to zero, so it converges by the alternating-series test. It is not absolutely convergent: the magnitudes are comparable to \(n^{-\alpha-1}\), where \(0<\alpha+1<1\), and the corresponding \(p\)-series diverges. If \(\alpha\leq-1\), the terms at \(x=1\) do not tend to zero, so the series diverges there.
At \(x=-1\), for negative \(\alpha\), the terms \((-1)^n\binom{\alpha}{n}\) are positive. If \(-1<\alpha<0\), their magnitudes tend to zero but are comparable to \(n^{-\alpha-1}\), whose exponent is between zero and one. Their positive sum diverges. If \(\alpha\leq-1\), the terms fail to tend to zero. Thus the series diverges at \(x=-1\) for every negative exponent.
| Exponent | At \(x=1\) | At \(x=-1\) |
|---|---|---|
| \(\alpha>0\) | Absolutely convergent | Absolutely convergent |
| \(\alpha=0\) | Terminates at \(1\) | Terminates at \(1\) |
| \(-1<\alpha<0\) | Conditionally convergent | Divergent; positive terms |
| \(\alpha\leq-1\) | Divergent; terms do not tend to zero | Divergent; terms do not tend to zero |
Whenever an endpoint series converges, its sum agrees with the limiting value from inside the interval of convergence. This follows from Abel's theorem: if a series \(\sum a_n\) converges to \(L\), then \(\sum a_n r^n\) tends to \(L\) as \(r\) increases to \(1\). Applying this at \(x=1\), or to the coefficients \((-1)^n\binom{\alpha}{n}\) at \(x=-1\), gives the endpoint values where the series converges. In particular, for \(-1<\alpha<0\) the conditionally convergent sum at \(x=1\) is \(2^\alpha\).
Worked Example: The Exponent Negative One Half
Let \(\alpha=-\frac12\). Applying the coefficient recurrence gives
Therefore
The radius is \(1\). At \(x=1\), the terms alternate, and their magnitudes decrease to zero; the series converges conditionally, with sum \(2^{-1/2}=1/\sqrt2\). At \(x=-1\), multiplying each coefficient by \((-1)^n\) makes every term positive. Their magnitudes are comparable to \(n^{-1/2}\), so the positive series diverges. In this case the terms do tend to zero, but that alone does not imply convergence. This checks the distinction between the two endpoints.
Worked Example: The Exponent Negative Two
For \(\alpha=-2\), the coefficient product simplifies directly:
The first terms are therefore
The coefficients are nonzero and their consecutive ratio in magnitude is \((n+2)/(n+1)\), which tends to \(1\); the radius is \(1\). At \(x=1\), the terms are \(1,-2,3,-4,\ldots\), which do not tend to zero, so the series diverges. At \(x=-1\), each term is \(n+1\), which also fails to tend to zero. Both endpoint classifications are therefore confirmed directly.
The Terminating Case and a Common Mistake
When \(\alpha=0\), the recurrence gives \(\binom{0}{0}=1\) and \(\binom{0}{n}=0\) for every \(n\geq1\). The series is exactly the constant polynomial \(1\), and it converges at both endpoints (indeed, for every real \(x\)). More generally, every nonnegative integer exponent gives a terminating polynomial with infinite radius of convergence.
A second endpoint error is to conclude that a series diverges merely because its terms are positive and tend to zero. For \(-1<\alpha<0\), the terms at \(x=-1\) do tend to zero, but their magnitudes behave like \(n^{-p}\) with \(0<p<1\); the positive sum still diverges. The relevant test must address the sum, not just the limit of its terms.
Summary
The Taylor coefficients of \((1+x)^\alpha\) are the generalized binomial coefficients. Unless \(\alpha\) is a nonnegative integer, the series has radius \(1\) and represents the function for \(|x|<1\). Positive exponents give absolute convergence at both endpoints; the zero exponent gives the constant polynomial; negative exponents give conditional convergence at \(x=1\) only when \(-1<\alpha<0\), and divergence at \(x=-1\) in every negative-exponent case.
Check Your Understanding
Use the coefficient recurrence and endpoint criteria to answer the following.
- Compute the coefficients through degree three for \((1+x)^{5/2}\).
- What is the radius of convergence when \(\alpha=-3/2\), and why does the ratio test not decide endpoint convergence?
- For \(\alpha=-1/3\), classify the series at each endpoint as absolutely convergent, conditionally convergent, or divergent.
- Why does the series for \(\alpha=0\) not follow the endpoint classification for nonterminating negative exponents?
- For \(\alpha=-1/2\), explain why the terms at \(x=-1\) tending to zero does not establish convergence.