# 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.

• Hint: $0\leq b_n\leq 12/n$ for $n\geq 3$. – sranthrop Sep 24 '17 at 23:08

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 $n>\max\left(3,\frac6\varepsilon\right)$.

• Do we really have $n+5<n$? ;) – sranthrop Sep 24 '17 at 23:13
• @sranthrop Thank you for the catch. I'm at a gym and working from a not-so-smart phone. – Mark Viola Sep 24 '17 at 23:17

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 $x \gt max(b+1,\, c)$.

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: $$n > \dfrac{1 + \sqrt{5}}{2} + \dfrac{24 - 2 \sqrt{5}}{\varepsilon} .$$