Show that a sequence does not have a limit How can I show that the sequence $x_n=\frac{(-1)^nn}{n+1}$ does not have a limit by using only the definition of limits?
Attempts:
Let's assume that our sequence has a limit $x$. Then there exists such a number $N$ so that for any $n>N$ $|a_n-a| < \epsilon$. This is also true for $\epsilon = \frac{1}{2}$. Now I take the numbers $N_1$, $2N_1$ and $2N_1+1$. But when I start subtracting (just like it is done with the sequence $x_n=(-1)^n$), I get a result which has $N_1$ in it, and thus I can not get a contradiction. What do I do now?

Intuition: The problem with this sequence is that it really should have two limits: $1$ and $-1$. (Consider separately what happens for odd $n$ and even $n$.)
Sketch of proof: To leverage our intuition into rigor, one way to proceed would be consider what the limit might be. If the sequence converges to $1$, then we have a problem because for odd $n$ the sequential values are negative and in particular at least one unit away from $1$. On the other hand, if the sequence converges to some number (call it $L$) that isn't $1$, then $L$ is separated from $1$ by some positive distance; this will be a problem when we consider the even subsequence, which will get arbitrarily close to $1$.
