Showing a sequence converges from the definition of limit I am having some trouble proving that this series converges to zero
$b_n=\dfrac{n+5}{n^2-n-1}$, $n\geq 2$
directly from the definition of limit. My attempts have only left me with a lengthy, complicated form for $n$ in terms on $\epsilon$ so I must be missing some simple step.
 A: Note from the estimates for $n\ge3$
$$n+5<3n$$
and 
$$n^2-n-1>\frac12n^2$$
we have immediately that for any given $\varepsilon>0$
$$0<b_n<\frac6n<\varepsilon$$
whenever $n>\max\left(3,\frac6\varepsilon\right)$.
A: In this approach we do not use the quadratic formula or any facts about the graphs of $2^{nd}$ degree polynomials.
By looking at the highest degree terms in the numerator and denominator in $\dfrac{n+5}{n^2-n-1}$, you know that as $n$ grows it will start 'acting like' $1/n$. So for all large $n$, 
$\tag 1 \dfrac{n+5}{n^2-n-1} \lt \dfrac{2}{n}$
It doesn't matter if you have doubts about (1), since we can easily verify that the following sequence (2)-(4) of statements are equivalent formulations of (1) when $n \ge 2$:
$\tag 2 n^2 + 5 n \lt \ 2 n^2 - 2 n - 2$
$\tag 3 n^2 - 7 n - 2 \gt 0$
$\tag 4  n^2 \gt 7 n + 2$ 
To verify (2) we must show that $n^2-n-1 \gt 0$ for $n \ge 2$; but $n^2-n-1 \ge 1$ since $n^2-n-2 = (n-2)(n+1)$. Perhaps the sequence was defined as starting at $n = 2$ to guarantee (in a humorous way) that not only is the denominator never equal to zero, but that $n^2-n-1 \gt 0$, allowing us to use cross multiplication on (1) and preserve the inequality.
So (1) is true if (4) is true.
We claim that (4) is true provided $n \ge 8$. To show this note first that $7 n + 2 = \frac{15}{2}(\frac{14}{15} n +\frac{4}{15})$ and that $n \ge \frac{14}{15} n +\frac{4}{15}$ if $n \ge 4$. By inspecting
$\tag 5 (n) (n) \gt (\frac{15}{2})(\frac{14}{15} n +\frac{4}{15})$
one can now easily verify the claim.
Now if $\varepsilon \gt 0$ is given, simply take $n \gt max(8,\frac{2}{\varepsilon})$.

Bonus exercise for pre-calculus students: 
Consider a $2^{nd}$ degree polynomial $f(x) = x^2 - b x - c$ with $b \gt 0$.
If  $\beta \gt b$ then $f(x) \gt 0$ whenever $x \gt max(\beta,\frac{c}{\beta - b})$. Letting $\beta = b + 1$, we get $f(x) \gt 0$ whenever $x \gt max(b+1,\, c)$.
A: Notice that $n^2 - n - 1 = \dfrac{1}{4} \left( 2n - 1 + \sqrt{5} \right) \left( 2n - 1 - \sqrt{5} \right)$ so we can rewritten:
\begin{align}
b_{n} & = \dfrac{4n+20}{\left( 2n - 1 + \sqrt{5} \right) \left( 2n - 1 - \sqrt{5} \right)} \\
& = \dfrac{4n - 2 + 2\sqrt{5} + 22 - 2\sqrt{5}}{\left( 2n - 1 + \sqrt{5} \right) \left( 2n - 1 - \sqrt{5} \right)} \\
& = \dfrac{2}{2n - 1 - \sqrt{5}} + \dfrac{22 - 2\sqrt{5}}{\left( 2n - 1 + \sqrt{5} \right) \left( 2n - 1 - \sqrt{5} \right)}
\end{align}
In addition, $22 - 2\sqrt{5} > 0$ and $2n - 1 + \sqrt{5} \geq 1, 2n - 1 - \sqrt{5} \geq 0, \forall n \geq 2$, we have further:
\begin{align}
\lvert b_{n} \rvert & \leq \left\lvert \dfrac{2}{2n - 1 - \sqrt{5}} \right\rvert + \left\lvert \dfrac{22 - 2\sqrt{5}}{\left( 2n - 1 + \sqrt{5} \right) \left( 2n - 1 - \sqrt{5} \right)} \right\rvert \\
& = \dfrac{2}{2n - 1 - \sqrt{5}} + \dfrac{22 - 2\sqrt{5}}{\left( 2n - 1 + \sqrt{5} \right) \left( 2n - 1 - \sqrt{5} \right)} \\
& \leq \dfrac{2}{2n - 1 - \sqrt{5}} + \dfrac{22 - 2\sqrt{5}}{2n - 1 - \sqrt{5}} = \dfrac{24 - 2\sqrt{5}}{2n - 1 - \sqrt{5}}
\end{align}
The last term less than $\varepsilon$ when:
$$ n > \dfrac{1 + \sqrt{5}}{2} + \dfrac{24 - 2 \sqrt{5}}{\varepsilon} . $$
