Generalized Binomial Series at Both Boundaries

Why the Two Boundary Points Need Separate Analysis

For a positive integer exponent, the binomial expansion of $(1+x)^\alpha$ ends after finitely many terms. For a noninteger exponent, it continues indefinitely. Its behavior at the edges of its interval of convergence is therefore a question about infinite series, not just algebraic expansion.

The two boundary points $x=1$ and $x=-1$ change the signs of the terms in different ways. We will track those signs and the size of the coefficients to decide whether each endpoint series converges absolutely or conditionally. In this case, both endpoint series converge absolutely, although their term signs look different.

The Generalized Binomial Series

Let $\alpha>0$ be noninteger. For each integer $n\geq 0$, define the generalized binomial coefficient by

\binom{\alpha}{0}=1, \qquad \binom{\alpha}{n} = \frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-n+1)}{n!} \quad (n\geq 1).

The generalized binomial theorem gives the expansion

(1+x)^\alpha = \sum_{n=0}^{\infty}\binom{\alpha}{n}x^n, \qquad |x|<1.

Since $\alpha$ is not an integer, none of the factors in the numerator becomes zero, so the expansion does not terminate. The question here is what happens when we substitute $x=1$ or $x=-1$. Those substitutions produce the endpoint series

\sum_{n=0}^{\infty}\binom{\alpha}{n} \qquad\text{and}\qquad \sum_{n=0}^{\infty}(-1)^n\binom{\alpha}{n}.

Definition: A series $\sum u_n$ converges absolutely if $\sum |u_n|$ converges. It converges conditionally if $\sum u_n$ converges but $\sum |u_n|$ diverges.

How the Coefficient Signs Change

Write $m=\lfloor\alpha\rfloor$, the integer satisfying $m<\alpha<m+1$. In $\binom{\alpha}{n}$, the factors $\alpha,\alpha-1,\ldots,\alpha-m$ are positive. Each subsequent factor is negative. Consequently, for $n\geq m+2$, there are $n-m-1$ negative factors, and

\operatorname{sign}\binom{\alpha}{n}=(-1)^{n-m-1} \qquad (n\geq m+2).

Thus the coefficients eventually alternate in sign. At $x=-1$, multiplication by $(-1)^n$ changes that alternating pattern: the endpoint terms $(-1)^n\binom{\alpha}{n}$ eventually all have the same sign, namely $(-1)^{m+1}$. This sign difference is worth noticing, but sign alone does not determine convergence. We must also measure the coefficient sizes.

Deriving the Coefficient Decay

Set $a_n=\left|\binom{\alpha}{n}\right|$. Once $n>\alpha$, the ratio of successive magnitudes is

\frac{a_{n+1}}{a_n} = \frac{n-\alpha}{n+1} = 1-\frac{\alpha+1}{n+1}.

This ratio gives a precise way to find the size of $a_n$. Choose an integer $N>\alpha$. Repeatedly applying the ratio formula gives

a_n = a_N\prod_{k=N}^{n-1} \left(1-\frac{\alpha+1}{k+1}\right), \qquad n>N.

Take logarithms. For sufficiently large $k$, the number $u=(\alpha+1)/(k+1)$ satisfies $0<u\leq 1/2$. The estimate $\log(1-u)=-u+O(u^2)$ then gives

\log a_n = \log a_N -(\alpha+1)\sum_{k=N}^{n-1}\frac{1}{k+1} +O\left(\sum_{k=N}^{n-1}\frac{1}{(k+1)^2}\right).

The sum of reciprocal squares remains bounded as $n$ grows, while the harmonic sum differs from $\log n$ by a bounded amount, as follows by integral comparison. Therefore

\log a_n=-(\alpha+1)\log n+O(1).

Exponentiating shows that there are constants $C_1,C_2>0$, depending on $\alpha$, such that for all sufficiently large $n$,

C_1n^{-\alpha-1} \leq \left|\binom{\alpha}{n}\right| \leq C_2n^{-\alpha-1}.

This two-sided comparison identifies the coefficient decay: its magnitude has the same convergence behavior as $n^{-\alpha-1}$. Since $\alpha>0$, we have $\alpha+1>1$, so the $p$-series $\sum n^{-\alpha-1}$ converges. Comparison now gives

\sum_{n=0}^{\infty}\left|\binom{\alpha}{n}\right|<\infty.

The Endpoint Theorem

Theorem: If $\alpha>0$ is noninteger, both endpoint series of the generalized binomial expansion converge absolutely: $$ \sum_{n=0}^{\infty}\binom{\alpha}{n} \qquad\text{and}\qquad \sum_{n=0}^{\infty}(-1)^n\binom{\alpha}{n}. $$ Their sums are $2^\alpha$ at $x=1$ and $0$ at $x=-1$, respectively. Neither endpoint series is conditionally convergent.

To justify the sums as well as the convergence, approach each endpoint from inside the interval. For $0<r<1$, the generalized binomial theorem gives

\sum_{n=0}^{\infty}\binom{\alpha}{n}r^n=(1+r)^\alpha, \qquad \sum_{n=0}^{\infty}\binom{\alpha}{n}(-r)^n=(1-r)^\alpha.

Because $\sum a_n$ converges, the terms in either series are bounded in magnitude by a single summable series: $\left|\binom{\alpha}{n}(\pm r)^n\right|\leq a_n$. To see why the endpoint limit can be taken term by term, first choose a tail whose sum of $a_n$'s is as small as desired. That tail bounds the corresponding tails for every $r\leq1$. For the remaining finite number of terms, $r^n$ tends to $1$ as $r$ tends to $1$. Thus the endpoint sums are the limits of the displayed interior sums. Taking $r$ upward to $1$ gives

\sum_{n=0}^{\infty}\binom{\alpha}{n} = \lim_{r\to1^-}(1+r)^\alpha = 2^\alpha,

and

\sum_{n=0}^{\infty}(-1)^n\binom{\alpha}{n} = \lim_{r\to1^-}(1-r)^\alpha = 0.

Absolute convergence at both endpoints follows because the absolute values of the terms are $a_n$ in either case. Since the absolute-value series converges, neither endpoint series is merely conditionally convergent.

Worked Examples at Both Endpoints

Worked Example: The Exponent One-Half

Take $\alpha=\frac12$. The coefficient recurrence is

\binom{\alpha}{n+1} = \binom{\alpha}{n}\frac{\alpha-n}{n+1}.

Starting with $\binom{1/2}{0}=1$, the first coefficients are

\binom{1/2}{1}=\frac12,\qquad \binom{1/2}{2}=\frac12\cdot\frac{-1/2}{2}=-\frac18, $$ $$ \binom{1/2}{3}=-\frac18\cdot\frac{-3/2}{3}=\frac1{16}, \qquad \binom{1/2}{4}=\frac1{16}\cdot\frac{-5/2}{4}=-\frac5{128}.

At $x=1$, the series begins $1+\frac12-\frac18+\frac1{16}-\frac5{128}+\cdots$. At $x=-1$, it begins $1-\frac12-\frac18-\frac1{16}-\frac5{128}+\cdots$. For $n\geq2$, the endpoint terms at $x=-1$ are negative, not alternating. In both cases their magnitudes are $\left|\binom{1/2}{n}\right|$, which are comparable to $n^{-3/2}$; the $p$-series test gives absolute convergence. The sums check by approaching the endpoints from inside: $(1+r)^{1/2}\to\sqrt2$ and $(1-r)^{1/2}\to0$. Thus the endpoint sums are $\sqrt2$ and $0$, respectively.

Worked Example: The Exponent Three-Halves

Take $\alpha=\frac32$. Applying the recurrence gives

\binom{3/2}{0}=1,\quad \binom{3/2}{1}=\frac32,\quad \binom{3/2}{2}=\frac32\cdot\frac{1/2}{2}=\frac38, $$ $$ \binom{3/2}{3}=\frac38\cdot\frac{-1/2}{3}=-\frac1{16},\quad \binom{3/2}{4}=-\frac1{16}\cdot\frac{-3/2}{4}=\frac3{128}, $$ $$ \binom{3/2}{5}=\frac3{128}\cdot\frac{-5/2}{5}=-\frac3{256}.

At $x=1$, the initial terms are $1+\frac32+\frac38-\frac1{16}+\frac3{128}-\frac3{256}+\cdots$. At $x=-1$, they are $1-\frac32+\frac38+\frac1{16}+\frac3{128}+\frac3{256}+\cdots$; the tail is positive. Since $\alpha+1=\frac52>1$, the coefficient magnitudes are comparable to $n^{-5/2}$, so both series converge absolutely. Their sums are $2^{3/2}=2\sqrt2$ and $0$: the interior values $(1+r)^{3/2}$ and $(1-r)^{3/2}$ tend to those numbers.

Worked Example: The Exponent Seven-Thirds

Now let $\alpha=\frac73$. The first coefficients are

\binom{7/3}{0}=1,\quad \binom{7/3}{1}=\frac73,\quad \binom{7/3}{2}=\frac73\cdot\frac{4/3}{2}=\frac{14}{9}, $$ $$ \binom{7/3}{3}=\frac{14}{9}\cdot\frac{1/3}{3}=\frac{14}{81}, \quad \binom{7/3}{4}=\frac{14}{81}\cdot\frac{-2/3}{4}=-\frac7{243}, $$ $$ \binom{7/3}{5}=-\frac7{243}\cdot\frac{-5/3}{5}=\frac7{729}.

At $x=1$, the coefficients first become negative at $n=4$ and alternate thereafter. At $x=-1$, the terms from $n=4$ onward are negative: the factor $(-1)^n$ turns the alternating coefficient signs into a fixed sign. Their magnitudes are comparable to $n^{-10/3}$, and $10/3>1$, so both endpoint series converge absolutely. The limiting interior values are $(1+r)^{7/3}\to2^{7/3}$ and $(1-r)^{7/3}\to0$, verifying the endpoint sums.

A Common Sign Mistake

A tempting but incorrect argument is that the $x=-1$ series must alternate, because substituting $-1$ introduces a factor $(-1)^n$. The signs of the coefficients must also be included. For $\alpha=\frac12$, for example, the terms begin

1,\quad -\frac12,\quad -\frac18,\quad -\frac1{16},\quad -\frac5{128},\quad \ldots

After the first two terms, this is a same-sign tail, not an alternating series. The correct convergence test is based on the magnitudes: their decay is comparable to $n^{-3/2}$, so the endpoint series converges absolutely. More generally, both endpoint series have the same absolute-value series, and that series converges for every $\alpha>0$.

Why the Endpoint Classification Matters

At a boundary point, a power-series radius does not by itself decide convergence; the endpoint series must be tested. Here, tracking the signs helps prevent incorrect claims about alternation, while the coefficient estimate provides the decisive convergence test. In particular, the positivity of $\alpha$ matters: it ensures that the decay exponent $\alpha+1$ exceeds $1$, giving absolute convergence at both endpoints.

Check Your Understanding

For each question, use the coefficient signs and the decay estimate rather than relying only on the appearance of a few terms.

For noninteger $\alpha>0$, what is the eventual sign pattern of $\binom{\alpha}{n}$, in terms of $m=\lfloor\alpha\rfloor$?
What is the eventual sign of $(-1)^n\binom{\alpha}{n}$, and why does it differ from the coefficient sign pattern?
What power of $n$ describes the size of $\left|\binom{\alpha}{n}\right|$ for large $n$?
Why do both endpoint series converge absolutely for every noninteger $\alpha>0$?
What are the sums at $x=1$ and $x=-1$, and how can they be verified using values inside the interval?