Any converging sequence is bounded 
Any converging sequence is bounded.

This is a theorem in our book. It is proved by picking the first (finite) number of elements and show that the others are smaller than some constant. But, I can't agree with this theorem. For example, $a_n=\frac{1}{n-1}$ is not bounded (by $n=1$) (or is it)?
Regards
 A: The definition of a convergent sequence is a sequence $a_n$ which has a limit $a$. We also know that:
$$\lim_{n\to\infty} a_n=a\iff\forall\varepsilon>0\exists N\in\mathbb N\forall n>N\colon|a_n-a|<\varepsilon$$
Note that as the comments indicate, we can always replace the first $k$ elements, for any given $k$, since we only care about what happens from some point. 
Taking a limit is what happens after the fat lady sings, in some sense. It is an asymptotic behavior of a sequence, it occurs when the sequences eventually moves towards a certain point.
From the above definition it should be clear why every convergent sequence is bounded. For only finitely many elements are bigger than $a+\dfrac{1}{2}$, where $a$ is the limit of the sequence.
A: When you use a induction, you need follow the sentence below:
$(\forall P)[[P(0) \land ( \forall k \in \mathbb{N}) (P(k) \Rightarrow P(k+1))] \Rightarrow ( \forall n \in \mathbb{N} ) [ P(n) ]]$
In other words, you need prove to $P(0)$ and prove if works to $P(k)$ works to $P(k+1)$ too.
In your example, $P(k) \Rightarrow P(k+1)$ holds, but you $P(0)$ no, because, $a_1=\frac{1}{1-1}$ don't exists.
But for a convergence sequence (note $a_1$,$a_2$,$a_3$,..., isn't a convergent sequence, because, $a_1$ isn't not greater than $a_2$, $a_1$ is indefinite) that $P(0)$ and $P(k) \Rightarrow P(k+1)$ holds the theorem are ok.
Note too, $a_2,a_3,a_4,...$ is a convergence sequence and bounded.
PS.: A good propostion P to this case is ask, is P(k) limitated?
