The series is convergent.
Explanation:[tex] \lim_{n \to \infty} \frac{1}{ 2^{n} } =[/tex] 1/∞ = 0 ( it would be divergent if the answer was: -∞ or +∞ )
Sum of the infinite geometric series:
S∞ = a1 / ( 1-r )
a1 = 1 a2=[tex] \frac{1}{ \sqrt{2} } [/tex] , a3=1/2, a4=[tex] \frac{1}{2 \sqrt{2} } [/tex]...
r =[tex] \frac{1}{ \sqrt{2} } [/tex]
S∞ =
