Show $\{x_n\}$ is cauchy if $x_n = \frac{3+n}{\beta + n}$ and $\beta >10$

Question

"Show $\{x_n\}$ is cauchy if $x_n = \frac{3+n}{\beta + n}$ and $\beta >10$"

I don't get the correct intuition behind $\beta >10$. I tried to solve the problem like this instead. $$\frac{3+n}{\beta +n}=\frac{n}{n+10}+\frac{3}{n+10}<n+\frac{3}{n}=\frac{n^2 +3}{n}$$

and $$\frac{n^2 +3}{n}<\frac{\epsilon}{2}$$

if $$n>\frac{2}{\epsilon}(n^2 +3).$$

If this were a valid option for $N$, we could then show $n,m > N \implies d(x_n, x_m)< \epsilon$. Why doesn't this method work? I think it's because this $N$ is too small.

Answer 1

First off, the Cauchy criteria applies for the difference of the terms $|a_m - a_n| < \epsilon$ and not for a single term. What you did is trying to solve for $a_n < \epsilon$ only. We have: for $m , n \ge N$, $|a_m - a_n| = \dfrac{(\beta - 3)|m-n|}{mn}< \dfrac{2(\beta-3)}{N}< \epsilon \iff N > \dfrac{2(\beta -3)}{\epsilon}$. Thus we can choose the $N$ based on the last inequality. Thus $\{a_n\}$ is Cauchy.

Answer 2

Hint: For given $\epsilon>0$, find $N$ such that $|x_n-1|<\frac 12\epsilon$ for all $n>N$. Then automatically $|x_n-x_m|<\epsilon$ for all $n,m.>N$

Answer 3

For cauchy convergence we must find how $|x_m - x_n|$ behaves. We have: $$|x_m - x_n| = |\frac{3+m}{b+m} - \frac{3+n}{b+m}| = |\frac{3 - b}{b+m} + 1 - 1 - \frac{3 - b}{b+n}| = |3 - b||\frac{1}{b+m} - \frac{1}{b+n}| \le |3 - b|(\frac{1}{b+m} + \frac{1}{b+n})\space\space (1)$$

Suppose ,without loss of generality, that $m > n$. Then: $$(1) \implies |x_m - x_n| \le 2\cdot|3 - b|\cdot\frac{1}{b+n}$$ For any $ε>0$, choose $n_0 = 2\cdot\frac{|3 - b|}{ε} - b$ and from here you can rove what you want.