I substituted a few natural numbers into $\frac{n}{n+1}$ but I still have no idea, how to solve this problem, I would appreciate any kind of help.
2026-04-19 04:26:21.1776572781
Notate in the decimal fraction form of $\frac{n}{n+1}$ the third digit after the decimal point with $a_n$. What are the limit points of $a_n$?
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Hint: Write $\dfrac{n}{n+1} = 1-\dfrac{1}{n+1}$. Then, for $n\ge1000$, we get $$ 1 > \frac{n}{n+1} = 1-\frac{1}{n+1} > 1 - \frac{1}{1000} = 0.999 $$