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

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.
