3
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.