# limit of a series homework question

I need some help with this question.

The question:

Prove or disprove the following statement:

If $$a_n\cdot a_{n+1} \rightarrow 0$$ and $$a_n > 0$$ for all $$n$$, then $$a_n \rightarrow 0$$

Solution attempt:

The solution I was given disproves this statement using an example (which I understand): $$a_n = \left\{\begin{matrix} 1\quad & {n\quad odd} \\ \frac{1}{n}\quad & n \quad even \end{matrix}\right.$$

but I don't understand what is wrong with this proof:

For all $$\epsilon>0$$, there exists an $$N$$ so that for all $$n\geq N$$, $$a_n\cdot a_{n+1} < \epsilon$$. Therefore (because $$a_n > 0$$ for all $$n$$):

$$a_n<\frac{\epsilon}{a_{n+1}}<\epsilon$$

which prooves the statemant.

Can anyone explain what I am doing wrong?

The inequality $$\frac{\epsilon}{a_{n+1}}<\epsilon$$ is not true. In fact, that inequality is only true if $$a_{n+1} > 1$$, which is not true for the particular example you cite. In fact, the inequality is reversed for all even values of $$n$$.