Probability of never getting k tails in infinite flips I am currently trying to solve a problem but am at a standstill. The question is:
If a coin is flipped infinitely many times, what is the probability that there will never be j successive tails?
I have a recursive sequence of a_n, where a_n is the number of ways of flipping a coin n times without having successive j tails. 
The recursive sequence is:
an=a(n-1) + a(n-2) + ... + a(n-j) (this is all based on the flips and what sequence the flips generate).
Now I am asked to prove that a_n < c^n for some c<2 for every n>=(j+1).
Not sure where to go from here.
 A: If you flip a coin with a nonzero probability of tails an infinite (either countable or uncountable, doesn't matter) number of times, then the probability of obtaining $j$ successive tails is always $1$ for $j \in \mathbb{N}$.
A: Suppose we flip a coin $j$ times, and the chance that any flip lands tails is $p \in (0,1)$.  Then, assuming that individual flips are independent, the probability that all flips land tails is simply $p^j$.  Let's call such a group of flips a single trial.  Consequently, if we repeat this experiment $n$ times--i.e., we conduct $n$ trials--the probability that none of the trials results in $j$ consecutive tails is $(1 - p^j)^n$.
Now, observe that the event that any $j$ consecutive tails, including those that may have started in one trial and ended in the next, is a strict superset of the aforementioned event, it follows that the probability in $N = nj$ flips of not observing $j$ consecutive tails is at least $(1 - p^j)^n$.  Taking the limit as $n \to \infty$, we find $$\lim_{n \to \infty} (1 - p^j)^n = 0,$$ since $0 < p < 1$ implies $0 < p^j < 1$, which implies $1 > 1 - p^j > 0$.  Therefore, the probability of avoiding eventually observing $j \ge 1$ consecutive flips--no matter how large $j$ is--is zero.
