Find radius of convergence for

$\sum _{n=1}^{\mathrm{\infty}}\frac{(x-1{)}^{n-3}+(x-1{)}^{n-1}}{{4}^{n}+{2}^{2n-1}}$

$\sum _{n=1}^{\mathrm{\infty}}\frac{(x-1{)}^{n-3}+(x-1{)}^{n-1}}{{4}^{n}+{2}^{2n-1}}$

studovnaem4z6
2022-05-16
Answered

Find radius of convergence for

$\sum _{n=1}^{\mathrm{\infty}}\frac{(x-1{)}^{n-3}+(x-1{)}^{n-1}}{{4}^{n}+{2}^{2n-1}}$

$\sum _{n=1}^{\mathrm{\infty}}\frac{(x-1{)}^{n-3}+(x-1{)}^{n-1}}{{4}^{n}+{2}^{2n-1}}$

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Calvin Oneill

Answered 2022-05-17
Author has **20** answers

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.

$|x-1|<4.$

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.

$|x-1|<4.$

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