<s>[INST] 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.
 [/INST] 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:
$$ n > \dfrac{1 + \sqrt{5}}{2} + \dfrac{24 - 2 \sqrt{5}}{\varepsilon} . $$
</s>