Showing sequence is Cauchy by Definition Question: I have to show that sequence $(x_n)$ defined by 

$x_n=\frac{n+(-1)^n}{2n-1}$ , $n=1,2,3,...$. 

Is Cauchy sequence using definition only. 

My attempt: (I can see given sequence is Cauchy because it is convergent sequence in $\mathbb{R}$)But I have to show it Cauchy by using definition of Cauchy sequence only. So I   considered 

$|x_n-x_m|= |\frac{n+(-1)^n}{2n-1}-\frac{m+(-1)^m}{2m-1}|$

$=|\frac{(n+(-1)^n)(2m-1)-(m+(-1)^m)(2n-1)}{(2n-1)(2m+1)}|$

$=|\frac{m-n+(-1)^n 2m+(-1)^{m+1}2n+(-1)^{n+1}+(-1)^{m+2}}{(2n-1)(2m-1)}|$

but I am unable to proceed further :-( please Help me...
Suppose $m&gt;n.$ $|x_m-x_n|=\left|\frac{m+(-1)^m}{2m-1}-\frac{n+(-1)^n}{2n-1}\right|\leq \left|\frac{m}{2m-1}-\frac{n}{2n-1}\right|+\frac{1}{2m-1}+\frac{1}{2n-1}=\frac{m-n}{(2m-1)(2n-1)}+\frac{1}{2m-1}+\frac{1}{2n-1}\leq \frac{3}{2n-1}&lt;\epsilon$ 

for sufficiently large $m,n.$
Using the work that you already have, you can bound your final equality by letting the -1 products be 1, i.e, $|\frac{m-n+(-1)^n 2m+(-1)^{m+1}2n+(-1)^{n+1}+(-1)^{m+2}}{(2n-1)(2m-1)}|$ &lt; $|\frac{m-n+2m+2n+1 + 1}{(2n-1)(2m-1)}|$. You can then expand the denominator and you'll have a cancellation.