How to prove the succession is unbounded OR the limit does not exist? $$a_n = (-1)^n + \dfrac{1}{n}$$
How can I prove the succession is unbounded? Or it has no limit?
By a first look, writing down the first terms, I inferred that as $n\to +\infty$ the series is oscillating between a number that is always closer to $1$ if $n$ is positive at the infinity, and to $-1$ if $n$ is negative at infinity.
So the succession is not regular, whence there is no limit.
How can I prove it with the formal definition? I tried with supposing the limit is $\ell$, and using the definition I need $\forall r > 0$ some $N\in\mathbb{N}$ such that for $n >N$ we have
$$\bigg| a_n - \ell \bigg| < r$$
Now here
$$\bigg| (-1)^n + \dfrac{1}{n}- \ell \bigg| < r $$
Yet from here I got stuck. How to deal with this? I don't know if I can use a general $\ell$ and to reach some absurd to prove there is no $\ell$.
But I don't know how to proceed in this case.
I thought I could use $|a+b| \leq |a| + |b|$ but I'm not sure if I can or it works.
 A: Cauchy method
You can simply use the fact that every convergent series is a Cauchy sequence. As your above sequence is not Cauchy, it cannot be convergent.
Why is it not Cauchy? Take any two consecutive elements $a_n$, $a_{n+1}$, wlog $n$ is even. Then you have
$$
|a_{n+1} - a_n| = |(-1)^{n+1}+\frac{1}{n+1}-(-1)^n-\frac{1}{n}|=|-2-\frac{1}{n}+\frac{1}{n+1}|
$$
Let $n$ be arbitrary then the two last terms will surely be $\leq 1$, so the absolute difference between two consecutive elements is always $\geq 1$. So for any $\epsilon < 1$ it is not possible to find a large enough index $N$ s.t. $|a_{n+1}-a_n|<\epsilon$ for all $n\geq N$, and thus $(a_n)_n$ is not Cauchy.
Elementar method
This is similar to the above method with a limit value in between:
Let's assume the limit is $l$ and take $\epsilon=0.5$ and assume there exists a $N$ s.t. $|a_n-l|<\epsilon$ for all $n\geq N$.
This would mean that $|a_{n+1}-l|<\epsilon$ as well. Combining these two results by using the triangle inequality gives:
$$
|a_{n+1}-a_n| = |(a_{n+1}-l) - (a_n-l)| \leq |a_{n+1}-l|+|a_n-l|<2\epsilon=1
$$
But if you insert the values for $a_n$ and $a_{n+1}$ on the left side as in the method above, the left side will always be $>1$ for any $n$. $\Rightarrow$ Contradiction, hence such an $N$ cannot exist and $a_n$ is not convergent.
