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Say you have two power series. One of them has ROC of 2, and the other one has an ROC of 4. If you add the two series together is the ROC ALWAYS the lesser ROC? It seems to be a trend I've noticed, not sure if it's a fact though.

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  • $\begingroup$ If the two ROC are different, the ROC of the sum is the smaller ROC. If the two ROC is the same, the ROC of the sum can become bigger. Look at answers for a related question for more details. $\endgroup$ – achille hui Nov 19 '13 at 9:08
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This isn't true. A silly example is to take any $f(x)$ at all with finite radius of convergence, and then $$ f(x) + (-f(x)) = 0 $$ so the right hand side has an infinite radius of convergence.

We can disguise this example as well as we like; e.g. Taylor expand something ugly like $e^x + \frac{\sin(x)}{x-3}$ around $x = 0$ and Taylor expand $\cos(x) - \frac{\sin(x)}{x-3}$ there too.

However, what you said will be true if the radii of convergence for the two functions are distinct. I think the easiest way to see this is complex-analytically, by interpreting the radius of convergence is equal to the distance to the nearest singularity. If $f$ is singular at $P$ and $g$ is not, then $f + g$ is singular.

I don't know if you've seen this (the complex-analytic property of the radius of convergence) before or not, but it is very nice for thinking about questions like yours. It doesn't work over the real numbers -- e.g. $\arctan(x)$ is perfectly continuous on all of $\mathbb{R}$ but the Taylor series at $x = 0$ only has radius of convergence 1; what's secretly going on is that $\arctan(x)$ blows up when $x$ gets close to $i$.

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