# What is the radius of convergence for the Taylor expansion of $\frac{e^x\sin(x)}{x^2+25}$ about $x=0$?

I would like to know how I could go about finding the radius of convergence for the Maclaurin series of $$\frac{e^x\sin(x)}{x^2+25}$$.

I am familiar with how to find the radius of convergence of more simple series such as the Maclaurin series of $$\frac{1}{1-x}$$, but I don't even know where I should start for something more complex like $$\frac{e^x\sin(x)}{x^2+25}$$.

$$\frac{e^x\sin x }{x^2+25}= e^x \cdot \sin x \cdot f(x)$$ where $$f(x) = \frac{1}{25} \frac{1}{1 + \left( \frac{x}{5} \right)^2}$$
Since $$e^x$$ and $$\sin x$$ have Maclaurin series with infinite radius of convergence, we need look only at $$f(x)$$ which has radius of convergence $$r=5$$.
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots, \quad (r=1)$$
$$\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \cdots, \quad (r=1)$$
$$\frac{1}{1+\left(\frac{x}{5}\right)^2} = 1 - \left(\frac{x}{5}\right)^2 + \left(\frac{x}{5}\right)^4 - \left(\frac{x}{5}\right)^6 + \cdots, \quad (r=5)$$
The radius of convergence of the product is determined by the intersection of the circles of convergence, which is a circle of radius 5 centered at the origin: $$r=5.$$