Show convergence of $a_n=\frac{n^3}{n^3+n}$

Question

I have $a_n=\frac{n^3}{n^3+n}$. I have to show it's convergent with limiting value 1. If we use our rules then we get: $$\lim_{n\to \infty}\frac{n^3}{n^3+n}=\lim_{n\to \infty}\frac{\frac{n^3}{n^3}}{\frac{n^3}{n^3}+\frac{n}{n^3}}=\lim_{n\to \infty}\frac{1}{1+\frac{n}{n^3}}=1.$$ But how do I show convergence here?

Answer 1

Hint : Prove that $a_{n} = \frac{n^{3}}{n^{3}+n} = \frac{1}{1+\frac{1}{n^{2}}}$ is increasing, which guarantees $a_{n}$ tends to $\sup\limits_{n \in \mathbb{N}} a_{n}$,since it's bounded, then show that it's supremum is 1.

Answer 2

$|a_n-1|=\frac n {n^{3}+n} <\frac 1 {n^{2}} <\epsilon$ if $n >\frac 1 {\sqrt {\epsilon}}$.

Answer 3

The function $f(x) = \frac{1}{1+x}$ is continuous at $0$ so $$ \lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{n^3}{n^3+n} = \lim_{n\to\infty} \frac{1}{1+\frac{1}{n^2}} = \frac{1}{1+\lim_{n\to\infty} \frac{1}{n^2}} = \frac{1}{1+0} = 1 . $$