Assuming ${a_n}$ is a convergent sequence, prove that the lim inf of $a_{n+1}$ is equal to the lim inf of $a_n$ I'm aware that you have to use the definition of a limit of a sequence, which is:
$\lim\limits_{n \to \infty} a_n = L $ if for every $E > 0$, there is an $N$ such that if $n >N $then $|a_n - L | < E$
I just have no idea how to combine the two different limits. 

First of all you should understand intuitively why this holds. The limit of a sequence $a_n$ tells us about the long-term behavior of the sequence. But $a_n$ and $a_{n+1}$ have the same long-term behavior, they are merely indexed differently; so they should have the same limit.
For the formal proof, suppose $a_n$ converges to $L$. Then like you say for any $\epsilon >0$, there is an $N$ such that if $n>N$, $|a_n-L|<\epsilon$. But in particular this tells us that if $n>N$, then $|a_{n+1}-L|<\epsilon$ as well (since $n+1>N$, so we may plug in $n+1$ into the previous inequality). This shows that $a_{n+1}$ converges to the same limit $L$.
