Convergence of the sequence $\frac{(-1)^{n+1}}{n}$ It can be easily shown that the sequence $(x_n) = \frac{1}{n}$ converges to 0. I tried to show that $\frac{(-1)^{n+1}}{n}$ also converges to 0 in a similar fashion.
My argument is as follows,
Since the sequence converges to 0, by definition for all positive $\epsilon$
$$
\Bigg|\frac{(-1)^{n+1}}{n}\Bigg| < \epsilon
$$
If $n$ is odd, then $\frac{1}{n} < \epsilon$, hence for any $\epsilon$ we can select $n > \frac{1}{\epsilon}$ and the inequality holds true. If $n$ is even, then $n > -\frac{1}{\epsilon}$ which trivially holds for all values of $n \; \text{and}\; \epsilon$
Is this correct ?
 A: 
If $n$ is even, then $n>−\frac 1ϵ$ which trivially holds for all values of $n$ and $ϵ$

That should be a big red flag that something is wrong.
You surely can't conclude that EVERY  value of $n$ will work trivially for EVERY value of $\epsilon$.  That would mean for instance that $-10^k \to 0$ because for EVERY $k$ and EVERY $\epsilon$ you have $-10^k < 0 < \epsilon$.
Surely something is wrong.
And what is wrong is you are taking absolute values.  $n > 0 > -\frac 1\epsilon$ will tell us that $\frac {(-1)^{n+1}}n  = \frac {-1}n <0< \epsilon$ which is trivially true for all even $n$ and all positive $\epsilon$ but it tells us absolutely nothing about whether $|\frac {(-1)^{n+1}}n|$ is greater or less than $\epsilon$.
And clearly if we take $n = 6$ and $\epsilon = \frac 1{10}$, knowing that $6 > -\frac 1{\epsilon} =-10$ and knowing that means $\frac {-1}{6} < \frac 1{10}$ does not mean that $|\frac {(-1)^7}6| < \frac 1{10}$.  That is surely false.
...
Instead.... As $|\frac{(-1)^n}n - 0| = |\frac 1n| = \frac 1n$, then for any $n > \frac 1{\epsilon}$ we will have $|\frac{(-1)^n}n - 0| = |\frac 1n| = \frac 1n< \epsilon$.
And that is all there is to it.
A: We say a sequence $x_n$ converges to 0, if for all $\epsilon>0$, there exists an N such that for all n>N, we have $|x_n|<e$.
In this case, $x_n=\frac{(-1)^{n+1}}{n}$, and therefore $|\frac{(-1)^{n+1}}{n}|=\frac{1}{n}$. Thus you can choose N large enough such that $\frac{1}{N}<\epsilon$, in which case you may conclude that the sequence converges to 0. I'm not too sure you need to consider the case where it's n is even or odd.
If you'd like a bit of a challenge you might want to prove that if ${x_{2m}}$ and ${x_{2m-1}}$  (ie even and odd sub sequences) converge to the same limit, then $x_n$ converges to that limit too, and try to relate this theorem to this particular sequence.
best of luck
