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I am trying to find the radius of convergence for the following function

$ f(x)=\sin(\pi x/4)$

I already found the Maclaurin series of the function and applied the ratio test but seems I cant get the radius of convergence right. I find the radius

$R = 4/\pi$

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    The radius of convergence is infinity, as it is always < 1, Thanks everybody for your help. Just found it.2012-04-23

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You have that the derivatives around $x=0$ of $y=\sin x$ are

$\{ y^{(n)}(0)\}=\{0,1,0,-1 ,\dots \}$

Since you have a multiplicative factor of $\dfrac{\pi }{4}$ this changes to

$\{ y^{(n)}(0)\}=\left\{0,\dfrac{\pi }{4},0,-\dfrac{\pi^3 }{4^3} ,\dots \right\}$

As a consequence you have that the coefficients are

$c_n=\frac{(-1)^n}{(2n+1)!} \left(\frac{\pi}{4}\right)^{2n+1}$

You can readily check that

$\lim \frac{c_{n+1}}{c_{n}}=0$

from which the radius is $\infty$, i.e., the whole extended real line.

As a general result, the convergence radius of the series for

$y=\sin(ax+b)$

is the whole real line.

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    @David Sorry, It was a typo. I'll fix it.2012-04-23