Solution
= Solution
{c}
The finite <geometric series> identity, obtained by subtracting $t$ times the sum from the sum itself, gives
$$
S_N=\sum_{n=0}^Nt^n=\frac{1-t^{N+1}}{1-t}.
$$
For $0<t<1$, the <geometric sequence> $t^{N+1}$ tends to zero by Question 2(a)(iii). Therefore
$$
\boxed{\sum_{n=0}^{\infty}t^n=\frac1{1-t}.}
$$