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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 $\frac{e^x\sin(x)}{x^2+25}$.

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$$\frac{e^x\sin x }{x^2+25}= e^x \cdot \sin x \cdot f(x)$$ 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: $r=5.$

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