gnatopoditw

2022-06-20

Determine the radius of convergence of the series $\sum _{n=1}^{\mathrm{\infty}}{a}_{n}{z}^{n}$ where ${a}_{n}=\frac{{n}^{2}}{{4}^{n}+3n}$

grcalia1

Beginner2022-06-21Added 23 answers

You have

${a}^{n}={4}^{-n}\frac{{n}^{2}}{1+3n{4}^{-n}},$

and therefore

$\underset{n\to \mathrm{\infty}}{lim}\sqrt[n]{{a}_{n}}=\underset{n\to \mathrm{\infty}}{lim}\frac{1}{4}\sqrt[n]{\frac{{n}^{2}}{1+3n{4}^{-n}}}=\underset{n\to \mathrm{\infty}}{lim}\frac{1}{4}\frac{{\sqrt[n]{n}}^{2}}{\sqrt[n]{1+3n{4}^{-n}}}=\frac{1}{4}.$

Therefore, the radius of convergence is 4

${a}^{n}={4}^{-n}\frac{{n}^{2}}{1+3n{4}^{-n}},$

and therefore

$\underset{n\to \mathrm{\infty}}{lim}\sqrt[n]{{a}_{n}}=\underset{n\to \mathrm{\infty}}{lim}\frac{1}{4}\sqrt[n]{\frac{{n}^{2}}{1+3n{4}^{-n}}}=\underset{n\to \mathrm{\infty}}{lim}\frac{1}{4}\frac{{\sqrt[n]{n}}^{2}}{\sqrt[n]{1+3n{4}^{-n}}}=\frac{1}{4}.$

Therefore, the radius of convergence is 4