So I'm having trouble proving that the sequence given by $(f_n) = ( \frac{n+3}{2n-1} )$ is Cauchy.
The solutions give the following hint:
"One possible $M$ is by solving $|f_n-f_m| \lt \frac{7m+7n}{mn} = \frac{7}{n} + \frac{7}{m} \lt \frac{\epsilon}{2} + \frac{\epsilon}{2}$. Then pick $M \gt \frac{14}{\epsilon}$."
I have zero clue how to do the steps in between $|f_n-f_m| \lt \frac{7m+7n}{mn}$, so any tips would be greatly appreciated.
Cheers!
Note that $$\frac{1}{2}<f_n =\frac{7}{4n-2} + \frac{1}{2} <\frac{2}{n} +\frac{1}{2}$$ for $n\geq 5$
For $\varepsilon$, there exists $M$ with $\frac{2}{M}<\varepsilon$ s.t. $n,\ m\geq M$ implies $$ 0< \frac{2}{n},\ \frac{2}{m}\leq \frac{2}{M}$$
Hence $\frac{1}{2}<f_n,f_m\leq \frac{1}{2} + \frac{2}{M}$ so that $$ |f_n-f_m| \leq \frac{2}{M}< \varepsilon$$