# Prove the sequence $\frac{n^2}{n^2-n-5}$ converges

I know the basic procedure for proving convergence of sequences, but I'm having difficulty reducing this sequence. I believe the sequence converges to 1, so I've set up the following two approaches:

1:

$\left|\frac{n^2}{n^2 - n - 5} - 1\right| = \left|\frac{n^2}{n^2 - n - 5} - \frac{n^2 - n - 5}{n^2 - n - 5}\right| = \frac{n + 5}{n^2 - n - 5}$

2:

$\left|\frac{n^2}{n^2-n-5} - 1\right| = \left|\frac{1}{1-\frac{1}{n}-\frac{5}{n^2}} - \frac{1-\frac{1}{n}-\frac{5}{n^2}}{1-\frac{1}{n}-\frac{5}{n^2}}\right| = \frac{\frac{1}{n}+\frac{5}{n^2}}{1-\frac{1}{n}-\frac{5}{n^2}}$

Both assume $n \geq 3$. Neither of these approaches appear to be leading me in the right direction. I can technically solve these for $f(n) < f(\epsilon)$, but both result in something very nasty, and I feel like I'm missing something simple.

You can do the formals by applying an estimate for the nominator and denominator. For $n>5$ we have that $0 < n+5 < 2n$ and $n^2 - n - 5 > n^2 - n^2/5 - n^2/5 = 3n^2/5$. This means that for $n>5$ we have
$$0<{n+5\over n^2-n-5}< {2n\over 3n^2/5} = {10\over 3}{1\over n}\to 0$$
Your second approach is also workable as the nominator $\to 0$ and the denominator $\to 1$ os $n\to \infty$. If you want/need to do $\delta$-$\epsilon$ you'd have to be a bit more elaborate: you estimate the denominator to be larger than a positive constant and then estimate the nominator with an factor of $1/n$.
For $n\to \infty$ we have $\frac{n^2}{n^2-n-5}=\frac{1}{1-\frac{1}{n}-\frac{5}{n^2}}$ (dividing numerator and denominator by $n^2$). All terms on the denominator go to 0 except 1 so the limit is 1.