Proving the reciprocal of a divergent sequence is convergent I want to prove that given a sequence (an) where the limit as n --> infinity of (an) = infinity, the limit as n --> infinity of 1/(an) = 0.
It's introductory real analysis, but I'm not sure where to start. Thanks
Edit: My thoughts:
If a sequence is divergent, given c>0 there exists an N s.t for all n>=N, (an) > c. 
Intuition tells me to use this definition (possibly negating it?) to prove what I'm looking for. Would that be correct?
 A: Suppose $\epsilon > 0$ is given.
Because $\lim_{n\to\infty} a_n = \infty$, there exists $N$ such that $n > N$ implies $a_n > \frac 1\epsilon$ $\implies$ $\frac{1}{a_n} < \epsilon$.
This is precisely what it means for $\frac{1}{a_n}$ to converge to $0$.
A: Okay, 
Suppose $a_n=+\infty$. Let $\epsilon > 0$ and let $M=1/\epsilon$.
Since $\lim a_n = +\infty$, there exist an $N$ such that $n>N$ implies $a_n>M=1/\epsilon$.
Therefore $n>N$ implies $\epsilon > 1/a_n > 0$ , Hence
$n>N \implies |1/a_n - 0| < \epsilon.$
By definition this means $\lim (1/a_n)=0$. 
You can prove this other way around as well and it is very similar. 
A: Reciprocal of a divergent (series), or a divergent (sequence)? There's huge difference. The series of 1/n diverges and so does the series of n.
I'm assuming you meant "sequence". In this case, this is also not necessary unless the sequence has a constant sign. The term "divergent" doesn't mean "goes to infinity". Finally, if you just want to prove that the reciprocal of a sequence of limit infinity is 0, then use the epsilon-delta definitions.
If (an) goes to infinity, then for any "large" M > 0, there's a positive integer k for which n > k => an > M.
This is the same as saying:
For any "small" e = 1/M > 0, there's k for which n > k => 1/an < 1/M = e.
This shows that 1/an converges to 0.
