promień zbieżności r szeregu potęgowego [tex]\sum_{n=0}^{\infty}a_{n}x^{n}[/tex] wyliczamy tak:
[tex]\lambda = \lim_{n \to +\infty} |\frac{a_{n+1}}{a_{n}}| = \lim_{n \to +\infty} (|a_{n}|)^{\frac{1}{n}}\\\\r=\frac{1}{\lambda}[/tex]
W tym szeregu [tex]a_{n} = (\frac{4n^{2}-2n+2}{5n^{2}-n+3})^{n}\\\lambda = lim_{n \to\infty} (|a_{n}|)^{\frac{1}{n}} = lim_{n \to \infty} ((\frac{4n^{2}-2n+2}{5n^{2}-n+3})^{n})^{\frac{1}{n}} = lim_{n \to \infty} (\frac{4n^{2}-2n+2}{5n^{2}-n+3}) = \frac{4}{5}\\[/tex]
więc
[tex]r = \frac{1}{\lambda} = \frac{1}{\frac{4}{5}} = \frac{5}{4}[/tex]
czyli szereg jest zbieżny gdy [tex]|x| < \frac{5}{4}[/tex]
