Taylor expansion of $(1+x)^α$ to binomial series – why does the remainder term converge?

Solution 1:

Again, I think that I got it. One needs to proceed differently:

Let $x ∈ ℝ$ such that $|x| < 1$. For $t ∈ ℝ$ between $x$ and zero, one has $0 ≤ |t| ≤ |x| < 1$ as well as $|x-t| < |1+t|$, because:

• for $x > 0$ one has $|x-t| = x - t < 1 < 1 + t = |1 + t|$, and
• for $x < 0$ one has $|x-t| = t - x < t + 1 = |1 + t|$, since $-x < 1$.

Therefore $\big|\frac{x-t}{1+t}\big|$ assumes a maximum $q<1$ for $t$ on the compact interval between zero and $x$. Then: \begin{align*} |R_n(x)| &= \Big| \int_0^x \frac{(x-t)^n}{n!} g_α^{(n+1)}(t) dt \Big|\\ &≤ \Bigg| \int_0^x \Big| \frac{(x-t)^n}{n!} g_α^{(n+1)}(t) \Big| dt \Bigg| \\ &\overset{(1)}{=} \Bigg| \int_0^x \Big| \frac{(x-t)^n}{(1+t)^n} ·(α-n) · \binom{α}{n} · (1+t)^{α-1} \Big| dt \Bigg|\\ &\overset{(2)}{≤} n · \binom{α}{n} ·|q|^n · \Bigg| \int_0^x |α/n -1| ·|1+t|^{α-1} dt\Bigg| \\ &\overset{(3)}{≤} n · \binom{α}{n} · |q|^n · (|α| +1) · C \overset{n → ∞}{\longrightarrow} 0, \end{align*} where the convergence follows from $|(n+1)\tbinom{α}{n+1}\big/n\tbinom{α}{n}|\overset{n → ∞}{\longrightarrow} 1$ and $|q| < 1$, and

• (1) $g_α^{(n)}(t) = (n+1)! \tbinom{α}{n+1}(1+t)^{α-n-1}$ and $\tbinom{α}{n+1} = \tfrac{α-n}{n+1} \tbinom{α}{n}$,
• (2) $\big|\frac{x-t}{1+t}\big|^n ≤ |q|^n$, and
• (3) $|α/n - 1| ≤ |α| + 1$, and $C := \Big| \int_0^x |1+t|^{α-1} dt\Big|$

Solution 2:

Let $f(z)=(1+z)^\alpha$ and $g(z)=\sum_{n\geq 0}\binom{\alpha}{n}z^n$.

If $\alpha$ is a nonnegative integer, it is clear that both functions coincide on $\mathbb{C}$ by the binomial theorem. So we assume that $\alpha\not\in\mathbb{N}$ from now on, so that $\binom{\alpha}{n}\neq 0$ for every $n\geq 0$.

Using the principal branch of the complex logarithm, we see that $g(z)=\exp(\alpha\log(1+z))$ is holomorphic on $\mathbb{C}\setminus (-\infty,-1]$, whence on the open unit disk $D$ in particular.

Now observe that $$ \frac{\Big|\binom{\alpha}{n+1}\Big||z|^{n+1}}{\Big|\binom{\alpha}{n}\Big||z|^{n}}=\frac{|\alpha-n||z|}{n+1}\longrightarrow |z|. $$ So the radius of convergence of $g$ is $1$ by the ratio test. Hence $g$ converges to a holomorphic function on $D$.

For every $x>0$, the remainder in the approximation of $f$ by its degree $n$ Taylor polynomial at $0$ is, by Taylor-Lagrange, $$ R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}=\binom{\alpha}{n+1}(1+c)^{\alpha-n-1}x^{n+1}=(1+c)^\alpha\binom{\alpha}{n+1}\left(\frac{x}{1+c} \right)^{n+1} $$ for some $0<c<x$. Since $0<\frac{x}{1+c}<1$, the convergence of $g$ at $\frac{x}{1+c}$ implies that the rhs above tends to $0$, since it is a constant times its general term. So the remainder of the Taylor approximation tends to $0$.

It follows that $f$ and $g$ coincide on $(0,1)$.

By the isolated zero principle, $f$ and $g$ coincide on the whole open unit disk $D$.