How to prove that $\frac{1+4n^2}{2+2n^2}$ is a cauchy sequence?

Question

I know that the following sequence is a

$$\frac{1 + 4n^2}{2+2n^2}$$

How can I show that it is a cauchy sequence - well it has to cause its convergent but I want to understand it with the cauchy definition.

I know that i have to get the $|a_m - a_n | < \varepsilon$ meaning that the distance between two elements has to be smaller than epsilon. But how big is epsilon? I only know that there is an element called $N_0$ when this will become true.

This is where my problem starts - what will be the next step?

So I have $ \frac{1+4n^2+1}{2+2n^2+1} - \frac{1+4n^2}{2+2n^2} < \varepsilon $ but I don't know how much $\varepsilon$ is nor do I know where $N_0$ is.

Answer 1

For $n \in \mathbb N$, you have: $$\frac{1 + 4n^2}{2+2n²}=\frac{4 + 4n^2 -3}{2+2n²}=2-\frac{3}{2+2n²}$$ Hence for $m \ge n$: $$\left\vert \frac{1 + 4n^2}{2+2n²}-\frac{1 + 4m^2}{2+2m²} \right\vert \le \frac{3}{2} \left(\frac{1}{1+n^2}+\frac{1}{1+m^2}\right) \le \frac{3}{1+n^2}$$ And for $n$ large enough, the RHS can be as small as desired.

Answer 2

Let $a_n$ the $n$-th term of the sequence, since $$a_n=\frac{1+4n^2}{2+2n^2}=\frac{(4+4n^2)-3}{2+2n^2}=2-\frac{3}{2}\frac{1}{n^2+1}$$ it follows $$|a_n-a_m|=\frac{3}{2}\left|\frac{1}{n^2+1}-\frac{1}{m^2+1}\right|\le\frac{3}{2}\left(\left|\frac{1}{n^2+1}\right|+\left|\frac{1}{m^2+1}\right|\right)\tag{1}$$ Given $\varepsilon>0$ by choosing $N_0$ such that $\frac{1}{1+N_0^2}<\varepsilon/3$ we get $$n,m >N_0\qquad\implies\qquad |a_n-a_m|\le 3\max\left(\left|\frac{1}{n^2+1}\right|,\left|\frac{1}{m^2+1}\right|\right)<3\frac{1}{1+N_0^2}<\varepsilon$$

Answer 3

For any $p > 1$, we have \begin{align} & \left|\frac{1 + 4(n + p)^2}{2 + 2(n + p)^2} - \frac{1 + 4n^2}{2 + 2n^2} \right|\\ = & \frac{2 + 2n^2 + 8(n + p)^2 + 8n^2(n + p)^2 - 2 - 2(n + p)^2- 8n^2 - 8n^2(n + p)^2}{4(1 + n^2)(1 + (n + p)^2)} \\ = & \frac{8p(2n + p) - 2p(2n + p)}{4(1 + n^2)(1 + (n + p)^2)} \\ = & \frac{3p(2n + p)}{2(1 + n^2)(1 + (n + p)^2)} \\ < & \frac{3p(2n + 2p)}{2n^2(n + p)(n + p)} \\ = & \frac{3}{n^2} \times \frac{p}{n + p} \times \frac{n + p}{n + p}\\ < & \frac{3}{n^2}. \end{align} From which you can see that the sequence is Cauchy.