Prove the sequence $f_{n} = \frac{1}{n^2+1}$ is a cauchy sequence.
I'm just making sure my logic and reasoning is sound for the above proof:
Definition of cauchy sequence: $f_n$ is Cauchy if for all $Ɛ$ there is an $M \in \mathbb{N}$ such that for all $n,m > M, |f_n-f_m|<Ɛ$
$|f_n-f_m|$=$|\frac{1}{n^2+1}-\frac{1}{m^2+1}|$
by the triangle inequality:
$\le|\frac{1}{n^2+1}|+|\frac{1}{m^2+1}|$
$<\frac{1}{n}+\frac{1}{m}$
To ensure $\frac{1}{n}+\frac{1}{m}<Ɛ$ it suffices to have:
$max(\frac{1}{n}+\frac{1}{m}) < \frac{Ɛ}{2} $, so we will take $M(Ɛ) = \lceil\frac{2}{Ɛ}\rceil$
Now for a formal proof:
Let $Ɛ>0$
Define $M(Ɛ) = \lceil\frac{2}{Ɛ}\rceil$
Let $n,m > M(Ɛ)$
$n,m > \frac{2}{Ɛ}$
$\frac{1}{n}<\frac{Ɛ}{2}$ and $\frac{1}{m}<\frac{Ɛ}{2}$
$|f_n - f_m| \le \frac{1}{n}+\frac{1}{m}$
$|f_n - f_m| < \frac{Ɛ}{2}+\frac{Ɛ}{2} = Ɛ$
Therefore $f_n$ is cauchy.
Solution to the proof
