Proving a sequence is unbounded The sequence is $(u_{n})_{n}$ for which $u_{n}=e^{n}$ 
The answer I have says that for any given positive real $U$, the term of index/position $n=\left \lfloor \ln(n+1) \right \rfloor + 1$  will be such that $\left | u_{n} \right |> U$. 
I don't understand this answer. Can someone please explain?
Thanks in advance.

Proving a sequence is unbounded means that you must show that for any large number $N$ there is no $M$ such that $u_M < N$ and that $u_k$ is also smaller than $N$ for every $k>M$.
For $u_k = e^k$ this can be done by showing that
$$e^k > N$$ if we pick $k = \lfloor \log (N+1)\rfloor + 1$
Because then you have
$$e^{\lfloor \log (N+1)\rfloor + 1} > N $$ since $N\le e^{\lfloor \log (N+1)\rfloor} \le N + 1$
Therefore for any bound $N$ you can show that the sequence outgrows it. 
For a sequence like $\{2,-1,20,-1,200,\ldots\}$ perhaps the easiest way to show it is unbounded is to forget about the part that doesn't fit the growing pattern. That is, consider instead the sequence $\{2, 20, 200\}$ You can show that this is unbounded, and it is a subsequence of the original sequence. Any sequence with an unbounded subsequence is itself bounded.
Hope that helps.
