Prove. $\frac{n}{n+1}$ is a Cauchy sequence.

Question

I'm a student studying for a test for my real analysis course. I've been given the above question. I'm going to attempt to solve it, if anyone has any tips or tricks to steer me in the right direction that would be much appreciated!

First I fix $\epsilon > 0$ and show for all $n,m >N, |a_n - a_m| < \epsilon $

$$\bigg| \frac{n}{n+1} - \frac{m}{m+1} \bigg| <\epsilon$$

$$\bigg| \frac{n(m+1)-m(n+1)}{(n+1)(m+1)} \bigg| <\epsilon$$

$$\bigg| \frac{n(m+1)-m(n+1)}{(n+1)(m+1)} \bigg| \leq |m+1| +|n+1| \leq |2m|+|2n| <\epsilon$$

$$2m<\frac{\epsilon}{2} \implies m < \frac{\epsilon}{4}$$

$$2n<\frac{\epsilon}{2} \implies n < \frac{\epsilon}{4}$$.

Therefore. $$|2m|+|2n| < \frac{\epsilon}{4} + \frac{\epsilon}{4} = \frac{\epsilon}{2}<\epsilon$$

Thanks for your time!

Answer 1

Your argument does not make sense. You did not come up with a suitable $N$.

$|\frac n {1+n}-\frac m {1+m}|= \frac {|n-m|} {(1+n)(1+m)} \leq \frac {|n-m|} {nm} |=|\frac 1n -\frac 1 m| \leq \frac 1n +\frac 1 m <\epsilon/2+\epsilon/2=\epsilon$ if $n >\frac 2 {\epsilon} $ and $m >\frac 2 {\epsilon} $. Take any $N> \frac 2 {\epsilon} $.

Answer 2

It converges to $1$. And every convergent sequence is Cauchy.

Answer 3

Given $\epsilon>0$.

for any $n,m$,

$$|a_m-a_n|=|a_m-1-(a_n-1)|$$ $$\le \frac{1}{ m+1}+\frac{1}{n+1}$$

Take $ N=\lfloor \frac{2}{\epsilon}\rfloor$.

then

$$n,m\ge N \;\;\implies \frac{1}{m+1}+\frac{1}{n+1}<2\frac{\epsilon}{2}$$

$$\implies \;\; |a_m-a_n|<\epsilon$$

$\implies (a_n) $ is Cauchy.