Answer is: option1
\( \frac{1}{8} \)Solution:
\[ \sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{2k} = \sum_{k=1}^{\infty}\left(\frac{1}{9}\right)^k \]
This is a geometric series with first term \( a=\frac{1}{9} \) and common ratio \( r=\frac{1}{9} \).
It converges to \[ S= \frac{\frac{1}{9}}{1-\frac{1}{9}} = \frac{1}{9-1} = \frac{1}{8} \]
