Proving the implication $\lim_{n\to\infty} a_n a_{n+1} =0 \Rightarrow \lim_{n\to\infty} a_n=0$ Let $\{a_n\}$ be a sequence such that for all $n$, $a_n>0$. I have to prove that if $\lim_{n\to\infty} a_n a_{n+1} =0$ then $\lim_{n\to\infty} a_n=0$. Here's my reasoning: $\forall\varepsilon>0,\exists k$ such that $\forall n>k$ we have $|a_n a_{n+1}|=a_n a_{n+1} < \varepsilon$ and thus $a_n=|a_n|=|a_n-0| < \frac{\varepsilon}{a_{n+1}} \Rightarrow \lim_{n\to\infty} a_n=0$. However I don't feel comfortable with the $a_{n+1}$ on the r.h.s. What do you think?

As others have said, this property doesn't hold. The flaw in your reasoning is the inequality that ${\varepsilon \over a_{n+1}}<\varepsilon$, you're assuming $a_{n+1}>1$, which you don't necessarily know. You can have $a_k<1$ for infinitely many $k$, which is what goes wrong with the examples given by other posters. 
Added:
The definition of limit says: $\left ( \forall \varepsilon ' >0 \right )\left (  \exists N \right )\left ( \forall n>N \right )\left (  |a_n -L|<\varepsilon \right )$. 
Let $\varepsilon '>0$ be given. If you fix some natural number $k$, you can set $\varepsilon = \left \lfloor\varepsilon ' |a_k| \right \rfloor$ to obtain an $N$ such that for all $n>N$: $|a_na_{n+1}|<\varepsilon$ which gives $|a_n|<{\varepsilon \over |a_{n+1}|}$. But notice that we have the fixed subindex $k$ in the numerator and $n+1$ in the denominator, so we cannot conclude anything further.
What you cannot do is set $\varepsilon=\left \lfloor |a_n| \varepsilon' \right \rfloor$ for all $n$ since this isn't a number, it is a function of $n$. And if you were to try to work around this by taking, for example $\varepsilon =\displaystyle \inf _{n \in \mathbb{N}}\left \lfloor |a_n| \varepsilon' \right \rfloor$ you have no guarantee that this is a positive number. 
