## Calculus: Early Transcendentals (2nd Edition)

$a_1 = 0.9$ $a_2 = 0.99$ $a_3 = 0.999$ $a_4 = 0.9999$ Converges. The limit of the sequence seems to be $1$.
$a_n = 1-10^{-n}$ $a_1 = 1 - 10^{-1} = 0.9$ $a_2 = 1 - 10^{-2} = 0.99$ $a_3 = 1 - 10^{-3} = 0.999$ $a_4 = 1 - 10^{-4} = 0.9999$ Converges and approaches $1$. The term $10^{-n}$ seems to approach zero as $n$ increases and $1-0=1$. Alternatively, simply looking at the terms tells us that the term approaches $1$ as $n$ increases.