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Find the radius and interval of convergence of the power series
$$ x-\dfrac{x^3}{3\cdot 3!} + \dfrac{x^5}{5\cdot 5!}+...+(-1)^{n+1}\dfrac{x^{2n-1}}{(2n-1)\cdot(2n-1)!}+.... $$
and then evaluate \(f'\left(\frac{\pi}{2}\right)\), where \(f\) is the function defined by the above series.

We have the series \(\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^{2k-1}}{(2k-1)(2k-1)!}\), now it is obvious that \(\displaystyle \limsup_{n\to\infty}\left|\frac{(-1)^{n+1}}{(2n-1)(2n-1)!}\right|=0\), so the radius of convergence is infinite and the interval we seek is \(\mathbb{R}\).

If \(\displaystyle f(x)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^{2k-1}}{(2k-1)(2k-1)!}\Rightarrow f'(x)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^{2k-2}}{(2k-1)!}=\frac{\sin(x)}{x}\), so \(\displaystyle f'\left(\frac{\pi}{2}\right)=\frac{2}{\pi}\).

Just to add that the evaluation of the derivative at \(\pi/2\) is allowed since within the circle of convergence we can determine the derivative by differentiating term by term.