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...
 A: Suppose $m>n.$ $|x_m-x_n|=\left|\frac{m+(-1)^m}{2m-1}-\frac{n+(-1)^n}{2n-1}\right|\leq \left|\frac{m}{2m-1}-\frac{n}{2n-1}\right|+\frac{1}{2m-1}+\frac{1}{2n-1}=\frac{m-n}{(2m-1)(2n-1)}+\frac{1}{2m-1}+\frac{1}{2n-1}\leq \frac{3}{2n-1}<\epsilon$ 
for sufficiently large $m,n.$
A: Using logic, algebra and simple inequality rules, one can show that for $n \ge 3$, $x_n$ is between the prior two terms, $x_{n-1}$ and $x_{n-2}$. So if $N \ge 3$ and $m \ge N$, the number $x_m$ lies inside the interval formed by $x_{N-1}$ and $x_{N-2}$. But then of course   the distance between $x_m$ and another term $x_n$ with $n \ge N$ can not be greater than the distance between $x_{N-1}$ and $x_{N-2}$.
If $N = 2k+1$, using the above reasoning, we know that $x_{N-2} \lt x_{N-1}$. Also
$\tag 1 x_{N-1} = \frac{2k + 1}{4k-1}$
and
$\tag 2 x_{N-2} = \frac{2k-2}{4k-3}$
Using algebra, we can write
$\tag 3 x_{N-1} -  x_{N-2} = \frac{8k-5}{(4k-1)(4k-3)} \; \text{ with } k \ge 1$
By making $\text{(3)}$ arbitrarily '$\varepsilon$ small', the assertion that $x_n$ is a Cauchy sequence can be backed up by finding an $N$ so that $|x_m - x_n| \le \varepsilon$ for $m,n \ge N$.
If it is your preference, you can make '$\le \varepsilon$' into '$\lt \varepsilon$' in at least two ways. But nothing changes - the two definitions of a Cauchy sequence, one with '$\le \varepsilon$' and the other using '$\lt \varepsilon$', are equivalent.
A: Using the work that you already have, you can bound your final equality by letting the -1 products be 1, i.e, $|\frac{m-n+(-1)^n 2m+(-1)^{m+1}2n+(-1)^{n+1}+(-1)^{m+2}}{(2n-1)(2m-1)}|$ < $|\frac{m-n+2m+2n+1 + 1}{(2n-1)(2m-1)}|$. You can then expand the denominator and you'll have a cancellation.
