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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 solutions

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    $\begingroup$ Yes, your proof is fine. $\endgroup$ – Mark May 4 at 8:53
  • $\begingroup$ Okay great, I was just checking as if you see in my question above, the solution was different. I thought my answer was too simple. $\endgroup$ – user11015000 May 4 at 8:55
  • $\begingroup$ Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not viewable to some, such as those who use screen readers. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 4 at 15:31
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    $\begingroup$ The proof before the red edit is fine. $\endgroup$ – DanielWainfleet May 4 at 18:52

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