Proof for the Binomial expansion for $(1+x)^n$

Question

At A level, is there a proof for the Binomial expansion for $(1+x)^n$ where $n$ is rational but not an integer, and $|x| < 1$, i.e., where $(1+x)^n = 1 + nx + (n(n-1)/2!)x^2 +\ldots$ or do we just have to accept it without any proof?

Answer 1

You can look at it as the same as your ol' expansion, just that binomial coefficients are replaced by their definitions because we define factorials of rationals differently. For example, $$\binom{n}{0}=1,\ \binom{n}{1}=n,\ \binom{n}{2}=\dfrac{n(n-1)}{2!},\ \cdots$$This might help in remembering the formula, but as said already, a proof is beyond your scope.

You can satisfy your curiosity by actually learning around some of these concepts, that would take a lot of time and devotion though, presuming you are a high school student.

Have a good day. :)

Answer 2

As mentioned in the comments, this proof is based on page 82 of an A-level text from 1932: Advanced Algebra Vol.1 : Durell, Clement V.

Let $\alpha$ be any real number, and $x$ any real number such that $|x| < 1.$ Consider the infinite series $$ \sum_{n=0}^\infty\binom{\alpha}nx^n, $$ where $$ \binom{\alpha}n = \frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}. $$ (The empty product expression for $n = 0$ evaluates to $1.$) Note the identity: \begin{equation} \label{4259137:eq:1}\tag{1} (n + 1)\binom{\alpha}{n+1} = (\alpha - n)\binom{\alpha}n, \end{equation} valid for all $n.$

For each $n \geqslant 1,$ let $f_n(x)$ be the sum of the first $n$ terms of that infinite series: $$ f_n(x) = \sum_{r=0}^{n-1}\binom{\alpha}rx^r. $$ Differentiating: $$ f_n'(x) = \sum_{r=1}^{n-1}r\binom{\alpha}rx^{r-1} = \alpha + \sum_{r=1}^{n-2}(r+1)\binom{\alpha}{r+1}x^r, $$ whence: \begin{align*} & \phantom{{}={}} (1 + x)f_n'(x) - \alpha f_n(x) \\ & = \sum_{r=1}^{n-2}\left[(r+1)\binom{\alpha}{r+1} + (r-\alpha)\binom{\alpha}r\right]x^r + (n-1-\alpha)\binom{\alpha}{n-1}x^{n-1} \\ & = -n\binom{\alpha}nx^{n-1}, \end{align*} by \eqref{4259137:eq:1}. Therefore: $$ \frac{d}{dx}\left[\frac{f_n(x)}{(1 + x)^\alpha}\right] = -\frac{n\binom{\alpha}nx^{n-1}}{(1 + x)^{\alpha+1}} \qquad (x > -1). $$ Integrating (note that the integral expression makes sense even when $x \leqslant 0$): \begin{equation} \label{4259137:eq:2}\tag{2} \frac{f_n(x)}{(1 + x)^\alpha} - 1 = -\binom{\alpha}n\int_0^x\frac{nt^{n-1}}{(1+t)^{\alpha+1}}\,dt. \end{equation} If we can show, for all $x$ such that $|x| < 1,$ that the expression on the right hand side of \eqref{4259137:eq:2} tends to $0$ as $n$ tends to infinity, then it will follow that $$ \lim_{n\to\infty}\frac{f_n(x)}{(1 + x)^\alpha} = 1. $$ That is, $\lim_{n\to\infty}f_n(x) = (1 + x)^\alpha,$ and this means: $$ \sum_{n=0}^\infty\binom{\alpha}nx^n = (1 + x)^\alpha \qquad (|x| < 1). $$

The denominator of the integrand in \eqref{4259137:eq:2} is always positive (for all values of $t$ in the interval $[0, x]$ if $x$ is positive, or the interval $[x, 0]$ if $x$ is negative). The sign of the numerator is constant; it is positive for all $t$ if $x$ is positive or if $n$ is odd, and it is negative for all $t$ if $x$ is negative and if $n$ is even. Therefore the absolute value of the integral is increased, or at least it is not decreased, if we replace the integrand by another integrand of greater absolute value. Since the denominator has its smallest value when $x$ is negative and $t = x,$ we certainly have: $$ \left\lvert\int_0^x\frac{nt^{n-1}}{(1+t)^{\alpha+1}}\,dt\right\rvert \leqslant \int_0^{|x|}\frac{nt^{n-1}}{(1-|x|)^{\alpha+1}}\,dt = \frac{|x|^n}{(1-|x|)^{\alpha+1}}. $$ (The denominator of this expression can be omitted if $x > 0,$ but it doesn't matter, because we are only interested in the behaviour of the expression as $n$ tends to infinity.) Therefore, from \eqref{4259137:eq:2}: \begin{equation} \label{4259137:eq:3}\tag{3} (1-|x|)^{\alpha+1} \left\lvert\frac{f_n(x)}{(1 + x)^\alpha} - 1\right\rvert \leqslant \left\lvert\binom{\alpha}n\right\rvert|x|^n. \end{equation} But for all $n > \alpha,$ we have (ignoring the trivial case $x=0$): $$ \frac{\left\lvert\binom{\alpha}{n+1}\right\rvert|x|^{n+1}} {\left\lvert\binom{\alpha}n\right\rvert|x|^n} = \frac{n - \alpha}{n + 1}|x| = \left(1 - \frac{\alpha + 1}{n + 1}\right)|x| \to |x| \text{ as } n \to \infty. $$ Therefore, for all sufficiently large values of $n,$ the ratio of successive terms of the sequence on the right hand side of \eqref{4259137:eq:3} - of course, this is just the $n^\text{th}$ term of the original infinite series, with $|x|$ in place of $x$ - is strictly less than $1.$ If we denote the sequence by $(u_n),$ then there exists a number $a$ such that $|x| < a < 1$ and a positive integer $N$ such that $u_n < u_Na^{n - N}$ for all $n > N.$ It would not be difficult to give exact values for $a$ and $N$ in terms of $\alpha$ and $|x|,$ but it doesn't matter. The important thing is that $a^{n - N} \to 0$ as $n \to \infty,$ therefore $u_n \to 0$ as $n \to \infty.$ This completes the proof.

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(This part is optional!)

Although the argument above is quite short and direct, deriving the sum of the infinite series almost from first principles, the part where it uses integration is quite obscure. (I found it obscure in the original version, and it is still obscure in my version.) It might be better, therefore, to avoid the use of integration altogether. The theorem on the existence of the integral of a general continuous function on a bounded closed interval is not needed, because all the integrands involved have known primitives, i.e., antiderivatives. All we need is the theorem that a function whose derivative is positive on an interval is increasing on that interval. (This is a consequence of the Mean Value Theorem.)

Given $x$ such that $|x| < 1,$ define, for all $t > -1$: \begin{align*} F(t) & = \frac{f_n(t)}{(1 + t)^\alpha} - 1, \\ G(t) & = \frac{-\binom{\alpha}nt^n}{(1-|x|)^{\alpha+1}}. \end{align*} (As before: we could omit the denominator of $G(t)$ if $x > 0,$ but it does no harm to leave it in.) Then $F(0) = G(0) = 0,$ and, as shown above: \begin{align*} F'(t) & = \frac{-n\binom{\alpha}nt^{n-1}}{(1+t)^{\alpha+1}}, \\ G'(t) & = \frac{-n\binom{\alpha}nt^{n-1}}{(1-|x|)^{\alpha+1}}. \end{align*}

Assume, for simplicity, that $\alpha$ is not a positive integer or zero. (The finite case of the Binomial Theorem can be assumed to be already known.) Then $\binom{\alpha}n \ne 0$ for all integers $n \geqslant 0.$ If $x > 0,$ then for all $t \in (0, x]$ we have $$ 0 > \frac{F'(t)}{\binom{\alpha}n} > \frac{G'(t)}{\binom{\alpha}n}. $$ If $x < 0$ and $n$ is odd, then the same inequalities hold for all $t \in [x, 0).$ On the other hand, if $x < 0$ and $n$ is even, then for all $t \in [x, 0)$ we have $$ 0 < \frac{F'(t)}{\binom{\alpha}n} < \frac{G'(t)}{\binom{\alpha}n}. $$ In all cases, therefore, the derivatives of the functions $F,$ $G,$ and $G - F$ have the same constant sign for all $t$ between $0$ and $x.$ Regardless of whether that sign is positive or negative, the initial condition $F(0) = G(0) = 0$ implies that $|F(x)| \leqslant |G(x)|,$ which is inequality \eqref{4259137:eq:3} above; and the argument is completed as before.

If anything, though, I find this argument harder, not easier to follow than the one involving an integral expression, even though technically it is simpler.