Showing that the Maclaurin series of $(1+x)^{-\frac32}$ converges, to the funtion, for $|x|<1$

Question

I know that I have to use Lagrange error and prove it approaches $0$ as $n → ∞$ but I do not know what to put for the $(n+1)$-th derivative of the function.

Answer 1

Generalized bynomial theorem.

$(a+b)^n = a^n + na^{n-1} b + \frac {n(n-1)}{2} a^{n-1}b^2+\cdots$

$c_{k+1} = \frac {c_k(n-k)}{k+1}$

This works for rational exponents, too.

$(1+x)^{-\frac 32} = 1^{-\frac 32} - \frac 32 1^{-\frac 52}x + \frac 52\cdot\frac 32 \cdot \frac 12 1^{-\frac 72}x^2 - \frac 32\cdot\frac 52 \cdot \frac 12\cdot \frac 72\cdot \frac 13 1^{-\frac 92}x^3\cdots$

$1^{-\frac n2} = 1$

$(1+x)^{-\frac 32} = 1 - \frac 32 x + \frac {5\cdot 3}{(2^2)(2!)}x^2 - \frac {7\cdot 5\cdot 3}{(2^3)(3!)}x^3+\cdots$

$(2n+1)\cdots7\cdot 5\cdot 3 = \frac {(2n+1)!}{2^{n}n!}$

$c_n = \frac {(-1)^n(2n+1)!}{2^{2n}(n!)^2}$

The radius of convergence is the largest $x$ such that $\lim_\limits{n\to\infty}|\frac {c_{n+1}}{c_n} x|< 1$ (the ratio test)

$|\frac {c_{n+1}}{c_n}| = \frac {(2n+3)(2n+2)}{2^2(n+1)^2}\\ \lim_\limits{n\to\infty}|\frac {c_{n+1}}{c_n}| = 1 \\ |x|<1$