obliczyć promień zbierzności szeregu
: 14 kwie 2007, o 14:05
\(\displaystyle{ \sum_{n=1}^{\infty}(sin^2\frac{1}{n})x^n}\)
\(\displaystyle{ \lim_{n\to\infty} \frac{\sin^{2}\frac{1}{n+1}}{\sin^{2}\frac{1}{n}} =\\
=\lim_{n\to\infty} \frac{\sin^{2}\frac{1}{n+1}}{(\frac{1}{n + 1})^{2}}\cdot \frac{(\frac{1}{n})^{2}}{\sin^{2} \frac{1}{n}}\cdot \left(\frac{n + 1}{n}\right)^{2} = 1}\)
Stąd:
\(\displaystyle{ R = 1}\)