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?