partial limits of a series let $a_n$ be a series and +5, -5 partial limits of it.
as well note that $\left|a_{n+1}-a_{n}\right| \leq \frac{1}{n}$ for all $n$.
prove that 0 is also a partial limit of $a_n$.
Thank you.
 A: The idea is this:
Imagine a child draws the real line on the ground and then starts hopping around on it in such a way that again and again, they land very close to "5" and to "-5". They can do this without ever landing close to "0": for example, they can just hop back and forth between "5" and "-5".
But now imagine that as time goes on, the child gets tired and their hops become smaller and smaller (actually: arbitrarily small). Then if they still go back and forth between "5" and "-5", then on the way they'll eventually have to land on or very close to "0": once their hops get smaller than $\epsilon$, if you divide the interval from "5" to "-5" into boxes of size $\epsilon$, then on their way from the one to the other, they will have to land in every box, including the box containing "0", since their hops aren't large enough to jump over an entire box in one hop. And if you wait long enough, you can do this for arbitrarily small $\epsilon$.
Can you turn that little story into a formal proof? :-)
A: It might be easiest to do a proof by contradiction.. suppose $0$ is not a partial limit. Then there is an $\epsilon > 0$ such that $|a_n| > \epsilon$ for all $n$. So if $n > {1 \over \epsilon}$, $a_{n+1}$ must have the same sign as $a_n$ since $|a_{n+1} - a_n| \leq {1 \over n} < \epsilon$. As a result, all $a_n$ for $n > {1 \over \epsilon}$ have the same sign, preventing both a positive and negative number from being a partial limit.
