<s>[INST] 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.
 [/INST] 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
$$n,m >N_0\qquad\implies\qquad |a_n-a_m|\le 3\max\left(\left|\frac{1}{n^2+1}\right|,\left|\frac{1}{m^2+1}\right|\right)<3\frac{1}{1+N_0^2}<\varepsilon$$
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