Real Analysis: proof limit with sigma-N definition of convergence

Question

Given a sequence Sn = $ \frac {1}{n^2+n}$ we have to prove the sequence converges to zero using the $\epsilon$-n definition of convergence. What I had in mind is that since |$ \frac {1}{n^2+n}$-0|<$\epsilon$ then $ \frac {1}{n^2+n}$<0 as n is a natural number, then there must exist a N that is ≤ n, and thus > 1/$\epsilon$. Thus I can prove the sequence Sn converges to 0. But this proof seem too simple to me, and I keep thinking I'm missing steps, as N never gets used.

Answer 1

Hint: Use that for $n\geq 1$ we have $$\frac{1}{n^2+n}<\frac{1}{n}$$ So it is $$n>\frac{1}{\epsilon}$$