Lagrange Remainder for an general binomial series We started this somewhat in class, but I wanted some more explanation in a general scenario. Given a binomial series $$(1+x)^{a} \approx 1+\alpha x + \frac{\alpha(\alpha -1)}{2!}x^{2} + \frac{\alpha (\alpha -1)(\alpha -2)}{3!}x^{3} + ... +\frac{\alpha (\alpha -1) ...(\alpha -n+2)}{(n-1)!}x^{n-1}$$
What is the Lagrange Remainder? My teacher went up to n=2, but I'm curious to see the remainder of the general binomial series as a whole. Not too sure how to do it well though.
 A: If
$$ f(x)= \sum_{k=0}^n \frac{f^{(k)}(a)(x-a)^k}{k!} +R_n(x), $$
the Lagrange form of $R_n$ is given by
$$ R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}, $$
for some $\xi \in (a,x)$.
In the case we are considering, $a=0$ and $f(x) = (1-x)^{\alpha}$, so the Taylor expansion is
$$ (1+x)^{\alpha} = \sum_{k=0}^n \binom{\alpha}{k}x^k + R_n(x) $$
The $(n+1)$th derivative of $f$ is
$$ f^{(n+1)} = \alpha(\alpha-1)\dotsm (\alpha-n) (1+x)^{\alpha-n-1}, $$
so the remainder is
$$ R_n(x) = \frac{\alpha(\alpha-1)\dotsm (\alpha-n)}{(n+1)!}(1+\xi)^{\alpha-n-1}x^{n+1} = \binom{\alpha}{n+1} (1+\xi)^{\alpha-n-1}x^{n+1}. $$
A: I think that a more clear approach is given by considering the integral remainder. 
Assume $|x|<1, n\in\mathbb{N}$ and $\alpha\not\in\mathbb{Z}$. Integration by parts gives:
$$(1+x)^{\alpha} = 1+\alpha x+\frac{(\alpha)_2}{2!}x^2+\ldots+\frac{(\alpha)_n}{n!}x^n+\frac{(\alpha)_{n+1}}{n!}\int_{0}^{x}(1+t)^{\alpha-n-1}(x-t)^n\,dt$$
where the Pochhammer symbol $(\alpha)_k$ means $\alpha(\alpha-1)\cdot\ldots\cdot(\alpha-k+1)=\frac{\Gamma(\alpha+1)}{\Gamma(\alpha+1-k)}.$
To estimate the remainder boils down to estimating the last integral, that by setting $t=xu$ equals:
$$x^{n+1} \int_{0}^{1}(1+xu)^{\alpha-n-1}(1-u)^n\,du = x^{n+1}\int_{0}^{1}\left(1+x(1-u)\right)^{\alpha-n-1} u^n\,du.$$
