Prove that the sequence has no limit. Problem: Prove that the sequence with the general term $x_n=1/n$, when $n=2k-1$ and $x_n=n/(n+2)$, when $n=2k$, ($k$ being a positive integer) has no limit.  
My attempt: It can be easily seen that for all odd numbers $x_n$ tends to $0$ and for all even numbers $x_n$ tends to $1$. Now for any $\epsilon>0$ we have to find a positive integer $N>0$ such that for all $n>N$ $|{x_n-L}|<\epsilon$, where $L$ is the assumed limit of the sequence. Possible values of $L\in[0,1]$.
Suppose we consider an arbitrary positive integer $a$, then we can conclude that for a small enough value of $\epsilon$ the inequality $|{x_n-L}|<\epsilon$ will not hold as $x_{a+1}$ or $x_{a+2}$ will tend towards either $0$ or $1$ (Depending on whether $L>0.5$ or $L<0.5$). Thus the sequence $x_n$ has no limit. 
The author of my textbook writes as follows: 

I'd like to know whether my attempt is correct. If not, how does it differ from the solution proposed by the author? 
 A: I think your ideas are fine, but to make it rigorous, you need to find some $\epsilon$ for which $|a_n-L|\geq \epsilon$ for all $n$ large enough (just negating the statement of convergence).
There are only 2 valid candidates for $L$, $0$ and $1$. How about picking $\epsilon=1/2$?
Equivalently, you can use the cauchy criterion for convergent sequences again $\epsilon=1/2$ will do.
A: You and the author are saying probably the same thing. However, to show that this sequence diverges, you can explicitly take a value of $\epsilon=1/8$: if the sequence $x_n$ has a limit $a$, then for some $N$ we would have $|x_n-a|<1/8$ whenever $n\geq N$.  
Notice that this implies in particular that $$|x_n-x_{n+1}|=|x_n-a+a-x_{n+1}|\leq |x_n-a|+|x_{n+1}-a|\leq \frac{2}{8}=\frac{1}{4}.$$ 
However, if $k\geq 2$, then $x_{2k-1}=\dfrac{1}{2k-1}\leq \dfrac{1}{3}$ and $$x_{2k}=\dfrac{2k}{2k+2}=\dfrac{2k+2-2}{2k+2}=1-\dfrac{2}{2k+2}\geq \dfrac{2}{6}=1-\dfrac{1}{3}=\frac{2}{3}.$$
Thus, the distance between $x_{2k}$ and $x_{2k+1}$ is at least $\frac{1}{3}$, so $|x_{2k-1}-x_{2k}|\geq \frac{1}{3}>\frac{1}{4}$. 
This shows that the given sequence does not converge.
A: You can take the difference of two consecutive terms ( when n is even and when n is odd) in terms of n and take limit of the difference as n tends to Infinity. It will be 1.
And thus the sequence diverges.
