prove that $(-1)^{n}n$ does not converge prove that $(-1)^{n}n$ does not converge
I need to prove this using the $N_{\epsilon}$ definition of limit. 
So I started with suppose there exists a limit $L$ such that $|(-1)^{n}n-L|<\epsilon$ for $n>N_\epsilon$. 
Next, I have $(-1)^{n}n<\epsilon+L$, and then $(-1)^{n}n<1^{n}n=n<\epsilon+L$, then $n<\epsilon+L$. But then I don't how to proceed, can someone help?
 A: First of all, as I said in my comment above, when disproving convergence, you are free to choose an $\epsilon$ of your liking. You should, of course, make sure that it works, but there is no reason to work as generally as you have done here. After all, a sequence converges if for any $\epsilon$, something something... That means if there is even a single $\epsilon$ where this something, something fails, then you have successfully disproven convergence.
I will choose $\epsilon = \frac12$, because I want to.
So assume, for contradiction, the sequence does converge to some $L$. I have my $\epsilon$, and that means that the definition of convergence grants me an $N$ such that for any $n > N$ we have $|(-1)^nn - L| < \frac12$. Specifically, this also means that the distance between any two $(-1)^nn$ and $(-1)^mm$ for $n, m > N$ can at most be $2\cdot \frac12 = 1$, since they're both at most $\frac12$ away from $L$. But this cannot be, for instance because the distance between $(-1)^nn$ and $(-1)^{n+2}(n+2)$ is $2$, no matter how big you make $n$. Specifically, the distance between two such terms is still $2$ for $n > N$.
This means that for $\epsilon = \frac12$, there can be no $N$ such that something, something. Therefore the sequence does not converge.
A: If a limit L existed, then for any $\epsilon>0$, there would exist a $N$ s.t. $|a_n-L|<\epsilon$ for all $n \geq N$.
Let $\epsilon=1$ and consider any $L$ and suppose that there existed some $N$ s.t. $|a_n-L|<1$ and $n\geq N$ is odd. Then 
$$|a_{n+1}-L|=|a_{n+1}-a_n+a_n-L|=|(n+1)+n+a_n-L|\geq 1$$
so no such N can exist. 
A: $u_n=(-1)^{n}n$ does not converge because it admit two subsequences ${u_{2n}}$ and ${u_{2n+1}}$ with two different limits $+\infty$ and $-\infty$ respectively.
A: Does this solution make sense? 
This is a proof by contradiction.
Let $\epsilon = \frac{1}{2}$.
Let $n $ be even, such that $n=2k$ for some $k\in \mathbb{N}$:
Assume the sequence “tends to” $l$, so that $\epsilon > |l-a_n|=\epsilon > |l-2k|$ and $\epsilon > |l-a_{n+1}|=|l-(2k+1)|$
So $\frac{1}{2} > |l-a_n|=\frac{1}{2}> |l-2k|$ and $\frac{1}{2}> |l-a_{n+1}|=|l-(2k+1)|$.
$\frac{1}{2} > |l-a_n|=\frac{1}{2}> |l-2k| \implies -\frac{1}{2}> l-2k >\frac{1}{2} $ And for $k+1$: $-\frac{1}{2} > l-2k-1 > \frac{1}{2} \implies \frac{1}{2}> l-2k > \frac{3}{2} $.  So, such an $l$ could not have existed because $l$ cannot be in both of those inequalities.
