## Multivariable Calculus, 7th Edition

We have $\lim\limits_{n \to \infty}a_n = \lim\limits_{n \to \infty}\frac{n^3}{n^3+1}$. Dividing both the numerator and denominator by the highest term in the denominator, $n^3,$ we get that the limit equals $\lim\limits_{n \to \infty}\frac{n^3/n^3}{n^3/n^3+1/n^3}= \lim\limits_{n \to \infty}\frac{1}{1+1/n^3} =\frac{1}{1+0} =1.$ Therefore, $\lim\limits_{n \to \infty}a_n =1$ and the sequence $\{a_n\}$ converges.