 # Find radius of convergence for <munderover> &#x2211;<!-- ∑ --> <mrow class="MJX-TeXAtom- studovnaem4z6 2022-05-16 Answered
$\sum _{n=1}^{\mathrm{\infty }}\frac{\left(x-1{\right)}^{n-3}+\left(x-1{\right)}^{n-1}}{{4}^{n}+{2}^{2n-1}}$
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The way it's written, it's a weird series to ask for radius of convergence, as it has a singularity at $x=1$
Forgetting about that, Alex Becker's answer is spot on: to find the radius of convergence you take the limit over n, not x, and only of the coefficients: here it would be
$R=\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n}}{{a}_{n+1}}=\underset{n\to \mathrm{\infty }}{lim}\frac{{\left(\frac{1}{4}\right)}^{n}}{{\left(\frac{1}{4}\right)}^{n+1}}=4.$
Or, if you want to go for understanding instead of just methods, again as Alex said this is a geometric series so it will converge when $|x-1|/4<1$, i.e.
$|x-1|<4.$