Hallo,
\( \frac{2^{2n}-1}{3^{2n}} \) = \( \frac{2^{2n}}{3^{2n}} \) -\( \frac{1}{3^{2n}} \)= (\( \frac{2}{3} \))^(2n) -\( \frac{1}{3^{2n}} \)
----> beide Terme gehen gegen 0
\( \lim\limits_{n\to\infty} \) (\( \frac{2^{2n}}{3^{2n}} \) -\( \frac{1}{3^{2n}} \)) =0
