Exercício 4.3. Mostre que, para todo x \in(-1,1), tem-se
\frac{1}{\sqrt{1-x^{2}}}=1+\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdots(2 n-1)}{2 \cdot 4 \cdot 6 \cdots 2 n} x^{2 n}

Passo 1

O enunciado pede para mostrar que, para todo x\in(-1,1),
\frac{1}{\sqrt{1-x^{2}}} = 1 +\sum_{n=1}^{\infty} \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\,x^{2n}.
O conteúdo que vamos usar é o teorema da série binomial para expoentes gerais, e em particular o caso \alpha=-\tfrac12.

Passo 2

O teorema da série binomial diz que, para |u|<1 e qualquer \alpha\in\mathbb{R},
(1+u)^{\alpha} = \sum_{n=0}^{\infty} \binom{\alpha}{n}\,u^{n},
onde
\binom{\alpha}{n} = \frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-n+1)}{n!}.

Passo 3

Tomamos \alpha=-\tfrac12 e u=-x^{2}. Então
(1 - x^{2})^{-\tfrac12} = \sum_{n=0}^{\infty} \binom{-\tfrac12}{n} (-x^{2})^{n} = \sum_{n=0}^{\infty} \binom{-\tfrac12}{n} (-1)^{n}x^{2n}.

Passo 4

Calculamos a forma explícita de \binom{-\tfrac12}{n}(-1)^{n}. Primeiramente,
\binom{-\tfrac12}{n} = \frac{(-\tfrac12)(-\tfrac12-1)\cdots(-\tfrac12-n+1)}{n!} = \frac{(-1)^{n}\,(1\cdot3\cdot5\cdots(2n-1))}{2^{n}\,n!}.
Multiplicando por (-1)^{n} obtemos
\binom{-\tfrac12}{n}(-1)^{n} = \frac{1\cdot3\cdot5\cdots(2n-1)}{2^{n}\,n!}.

Passo 5

Logo a série fica
(1 - x^{2})^{-\tfrac12} = \sum_{n=0}^{\infty} \frac{1\cdot3\cdot5\cdots(2n-1)}{2^{n}\,n!}\,x^{2n}.
Para n=0 o fator vazio vale 1, dando o termo inicial 1. Para n\ge1, note que
2^{n}\,n! = 2\cdot4\cdot6\cdots(2n),
pois 2\cdot4\cdot\cdots(2n)=2^{n}(1\cdot2\cdot\cdots n)=2^{n}\,n!. Assim
\frac{1\cdot3\cdot5\cdots(2n-1)}{2^{n}\,n!} = \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}.

Passo 6

Concluímos que, para todo x\in(-1,1),
\frac{1}{\sqrt{1-x^{2}}} =(1 - x^{2})^{-\tfrac12} =1+\sum_{n=1}^{\infty}\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\,x^{2n}.

Resposta

\frac{1}{\sqrt{1-x^{2}}} = 1 +\sum_{n=1}^{\infty} \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}\,x^{2n}.

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