Show that $\frac{n}{n^2-3}$ converges

Question

Hi I need help with this epsilon delta proof. The subtraction in the denominator as well as being left with $n$ in multiple places is causing problems.

Answer 1

If $n \geq 3$ then $n^2-3>n^2/2$ hence the quotient is less than $2/n $. Can you finish from here?

The point of this initial step was to get rid of the subtraction by absorbing it into a constant factor. This is desirable because, as you noticed, the subtraction in the denominator makes it harder to get a good upper bound. You would do the opposite to absorb a problematic addition in the numerator.

Answer 2

@Ian has provided a very efficient way to tackle a problem such as this. I thought it might be instructive to see how tight a bound we can achieve. To that end, let's write for $n\ge2$

$$\begin{align} \left|\frac{n}{n^2-3}\right|<\epsilon \implies n&>\frac{1}{2\epsilon}+\frac{1}{2\epsilon}(1+12\epsilon^2)^{1/2} \tag 1\\\\ &=\frac{1}{2\epsilon}+\frac{1}{2\epsilon}\left(1+6\epsilon^2+O\left(\epsilon^4\right)\right)\\\\ &=\frac{1}{\epsilon}+3\epsilon+O\left(\epsilon^3\right) \end{align}$$

where $(1)$ provides the tightest bound for $N$ as $N=\left\lceil \left(\frac{1}{2\epsilon}+\frac{1}{2\epsilon}(1+12\epsilon^2)^{1/2} \right)\right \rceil$.

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Now, we could also proceed as follows. We note that $n^2-3>\frac{n^2}{1+\alpha}$, for any $\alpha>0$, provided that $n\ge \left \lceil \sqrt{3\frac{1+\alpha}{\alpha}}\right \rceil$. For $\alpha =1$, $n^2-3>\frac12 n^2$ for $n\ge \left \lceil \sqrt{6} \right \rceil =3$, which is the inequality imposed by @Ian.

So, given $\epsilon>0$, we could actually take $N=\max\left(\frac{1+\alpha}{\epsilon},\left \lceil \sqrt{3\frac{1+\alpha}{\alpha}}\right \rceil\right)$, for any $\alpha >0$ and

$$\left|\frac{n}{n^2-3}\right|<\epsilon$$

whenever $n>N$.

Answer 3

$$\left|\frac{n}{n^3 - 3} \right| < \epsilon \implies n > N$$

For some $N$.

Since $n > 1$ (we can set this because you are searching convergence), $$n^3 - 3 < n^3 \implies \frac{1}{n^3 - 3} > \frac{1}{n^3}$$

Giving,

$$\left| \frac{n}{n^3} \right| < \left|\frac{n}{n^3 - 3} \right| < \epsilon$$

$$\to \left| \frac{1}{n^2} \right| < \epsilon$$

Finally,

$$n^2 > \epsilon \implies n > \sqrt{\epsilon}$$

If you want to keep $N \in \mathbb{N}$ then take $N = \lfloor \sqrt{\epsilon} \rfloor$.