Show $(-1)^n+\frac 1n$ diverges from the definition 
Show from the definition that $$s_n = (-1)^n + \frac 1n$$ is divergent.

I'd like to use subsequences here, but we haven't gotten to that yet (we just started sequences so we haven't even gotten to the fact the limit of a sum is the sum of the limits yet).  Besides, it says to use the definition.  But the definition
$$\forall \epsilon >0\ \exists N\in \Bbb N\ \forall n\ge N\ :\ |s_n - L| \lt \epsilon$$
has that $L$ in it.  How am I supposed to used this definition if there is no $L$?
 A: Hint: Assume it has a limit $L$. Clearly $L>0$ or $L<0$. Treat each case separately and demonstrate infinitely many  points that are $<0$ (if $L>0$) and vice versa for the second case.
This shows you can find an $\varepsilon>0$ (what is it exactly?) such that $|s_k -L|>\varepsilon$ for some $k>N$ for any $N$.
A: You're supposed to show that there is no possible $L$ such that the definition holds.  That is, for any $L$, you want to show that there exists an $\epsilon>0$ such that for any $N\in\mathbb{N}$, there exists $n\geq N$ such that $|s_n-L|\geq \epsilon$.
Let me show how you how to do this for one particular value of $L$.  Say $L=1$.  Then we can choose $\epsilon=1$.  Note that for any odd $n$, $s_n=(-1)^n+1/n=-1+1/n\leq 0$, so $|s_n-L|\geq 1$.  In particular, given any $N$, you can pick any odd $n\geq N$ and then $|s_n-L|\geq \epsilon$.
Now you want to generalize this argument to work no matter what $L$ is.  It's not so easy to give a single argument that works for every possible value of $L$ at once, so you may want to split into some different cases.  For instance, you might give one argument when $L\geq 0$, and a different argument when $L<0$.  You may also want to choose your $\epsilon$ to depend on $L$.
A: HINT: show the negative of the statement:
$$\exists\epsilon>0,\forall N\in\Bbb N,\exists n\ge N:|s_n-L|>\epsilon,\forall L\in\Bbb R$$
A: It will be useful to define the $N$-th tail of a sequence $(s_{n})_{n \geq 0}$ as the sequence
$$
s_{N+1}, s_{N+2}, \ldots,
$$
which is obtained by dropping the first $N$ terms from $(s_{n})_{n \geq 0}$.
Now, a sequence can be said to converge to $L$ if every neighborhood of $L$ contains some tail of the sequence.  (This definition of convergence is equivalent to the others.)
With your sequence, an $N$-th tail with $N$ big enough will have at least two terms that are at least distance $1/2$ apart (because one term is very close to $1$, and the other to $-1$).  Therefore, such a tail cannot be contained in any neighborhood (of any point) that has size smaller than $1/2$.  I.e., the sequence cannot be convergent.
(The above is related to the Cauchy criterion, which I avoided introducing.)
A: If you are allowed to use the fact that $a_n = (-1)^n$ takes only two values, $\{-1,1\}$, then, from the fact that the set of natural numbers is countable, it follows that $a_n$ takes both values infinite number of times, hence $L$ does not exist. Now show that $b_n = \frac{1}{n}$ converges to $0$ for $n >N$: this is easy by counter-example. Sum of convergent and divergent series diverges.  
