Calculus: Early Transcendentals (2nd Edition)

Converges to $10$
$a_n = \frac{100n-1}{10n}$ $a_1 = \frac{99}{10}$ $a_2 = \frac{199}{20}$ $a_3 = \frac{299}{30}$ $a_4 = \frac{399}{40}$ $a_5 = \frac{499}{50}$ $a_6 = \frac{599}{60}$ $a_7 = \frac{699}{70}$ $a_8 = \frac{799}{80}$ $a_9 = \frac{899}{90}$ $a_{10} = \frac{999}{100}$ The terms seem to get closer to the value $10$, meaning that the sequence is converging.