(a)
\( (1-x)^{-1/2} = \sum_{i=0}^\infty \binom{-1/2}{i}(-x)^i = 1 + \frac{1}{2}x + \frac{\frac{-1}{2}\frac{-3}{2}}{2}x^2 + \frac{\frac{-1}{2}\frac{-3}{2}\frac{-5}{2}}{3!}x^3 + \frac{\frac{-1}{2}\frac{-3}{2}\frac{-5}{2}\frac{-7}{2}}{4!}x^4 + \frac{\frac{-1}{2}\frac{-3}{2}\frac{-5}{2}\frac{-7}{2}\frac{-9}{2}}{5!}x^5 + \cdots = \)
\( = \sum_{i=0}^\infty \binom{-1/2}{i}(-x)^i = 1 + \frac{1}{2}x + \frac{\frac{3}{4}}{2}x^2 + \frac{\frac{-15}{8}}{3!}x^3 + \frac{\frac{105}{16}}{4!}x^4 + \frac{\frac{945}{32}}{5!}x^5 + \cdots = \)
\( = 1 + \frac{1}{2}x + \frac{3}{8}x^2 + \frac{5}{16}x^3 + \frac{35}{128}x^4 + \frac{63}{256}x^5 + \cdots \)
(b)
Utilizaremos a seguinte notação:
\( (1-x)^{-1/2} = \sum_{i=0}^\infty \binom{-1/2}{i}(-x)^i = \sum_{i=0}^\infty b_{i}x^{i} = \sum_{i=0}^\infty a_{i} \)
Considere agora:
\( \frac{\binom{\alpha}{i+1}}{\binom{\alpha}{i}} = \frac{\alpha(\alpha-1)...(\alpha-(i+1)+1)}{(i+1)!} \frac{i!}{\alpha(\alpha-1)...(\alpha-i+1)} = \frac{\alpha-i}{i+1} \)
Então para \( \alpha = \frac{-1}{2} \),
\( \frac{a_{i+1}}{a_i} = -(\frac{1/2+i}{i+1}), (-x)<x \)
Usando isso, descobrimos que:
\( b_{i+1}<b_i\frac{1}{50} \)
\( b_{i+2}<b_{i+1}\frac{1}{50}<b_i(\frac{1}{50})^2 \)
Para \( x = \frac{1}{50} \). Então, por indução, \( b_{n+i}<b_{n}(\frac{1}{50})^i \)
\( r_n = \sum_{i=1}^\infty a_{n+i} < \sum_{i=1}^\infty a_{n}(\frac{1}{50})^i = a_n \frac{1/50}{1-1/50} = \frac{a_n}{49} \)
\( r_n < \frac{a_n}{49} \)
(c)
Note que \( (1-x)^{-1/2} = (1-\frac{1}{50})^{-1/2} = (\frac{49}{50})^{-1/2} = \frac{5\sqrt{2}}{7} \)
\( (1-\frac{1}{50})^{-1/2} = 1 + \frac{1}{100} + \frac{3}{2}(\frac{1}{100})^2 + \frac{5}{2}(\frac{1}{100})^3 + \frac{35}{8}(\frac{1}{100})^4 + \frac{63}{8}(\frac{1}{100})^5 \approx 1,010152545 \)
\( \frac{7}{5} (1-\frac{1}{50})^{-1/2} \approx 1,414213563 \approx \sqrt{2} \)