[tex] \sum\limits_{k=1}^{\infty}420\left(\dfrac{1}{6}\right)^{k-1} [/tex]
The infinite geometric series is converges if |r| < 1.
We have r=1/6 < 1, therefore our infinite geometric series is converges.
The sum S of an infinite geometric series with |r| < 1 is given by the formula :
[tex]S=\dfrac{a_1}{1-r}[/tex]
We have:
[tex]a_1=420\left(\dfrac{1}{6}\right)^{1-1}=420\left(\dfrac{1}{6}\right)^0=420\\\\r=\dfrac{1}{6}[/tex]
substitute:
[tex]S=\dfrac{420}{1-\frac{1}{6}}=\dfrac{420}{\frac{5}{6}}=420\cdot\dfrac{6}{5}=84\cdot6=504[/tex]
Answer: d. Converges, 504.
