Let $$f(x)= x+\frac2{1\cdot3}x^3+\frac{2\cdot4}{1\cdot3\cdot5}x^5+\frac{2\cdot4\cdot6}{1\cdot3\cdot5\cdot7}x^7+\cdots\quad\forall x\in(0,1)$$ If the value of $f(\frac12)$ is $\dfrac\pi{a\sqrt b}$ (where $a, b\in\Bbb R$), then what is $|a+b|$?
2026-04-07 22:59:43.1775602783
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if $f(x) = x+2/(1\cdot3)x^3+2\cdot4/(1\cdot3\cdot5)x^5+\cdots$ comment on the value of $f(1/2)$
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With a series such as
$$ f(x)= x+\frac2{1\cdot3}x^3+\frac{2\cdot4}{1\cdot3\cdot5}x^5+\frac{2\cdot4\cdot6}{1\cdot3\cdot5\cdot7}x^7+\cdots $$
one approach is to expand it into a exponential generating function
$$ f(x) = x + \frac{2^2}{3!}x^3 + \frac{(2\cdot 4)^2}{5!}x^5 + \frac{(2\cdot 4\cdot 6)^2}{7!}x^7 + \cdots. $$
The numerators are $\, 1,\,4,\,64,\,2304,\,147456,\,\cdots.\,$ A lookup of this sequence in the OEIS finds that this is OEIS sequence A002454 where one of the formulas is
E.g.f.: ... = arcsin(x)/sqrt(1-x^2) = ...
This implies that
$$ f(1/2) = \frac{\sin^{-1}(1/2)}{\sqrt{1-1/4}} = \frac{\pi/6}{\sqrt{3}/2} = \frac{\pi}{3\sqrt{3}}. $$
$$f(x)=\frac{1}{2}\sum_{k=0}^{\infty} \frac{(2x)^{2k+1}}{(2k+1){2k\choose k}}=\frac{\sin^{-1}(x)}{\sqrt{1-x^2}}.$$ $$\implies f(1/2)=\frac{\pi}{3\sqrt{3}}.$$