Proving Divergence of |cos(n)a_n| w/ Converging Series

[daniel_i_l]

Homework Statement

Lets say that I have some sequence $$(a_n)$$ which converges to 0 at infinity and that for all n $$a_{n+1} < a_n$$ but the sequence $$(a_n)$$ diverges. Now I know that the series
$$(cos(n) a_n)$$ converges but can I use the following argument to prove that
$$|cos(n) a_n|$$ doesn't converge:
$$|cos(n) a_n| >= {cos}^{2}(n) a_n = {a_n}/2 + {(cos(2n)) a_n}/2$$
And since $${(cos(2n)) a_n}/2$$ converges and $${a_n}/2$$ diverges
$${cos}^{2}(n) a_n$$ diverges and so $$|cos(n) a_n|$$ diverges.
Is that always true?
Thanks.

 

[Dick]

That seems ok, provided you can prove cos(2n)*a_n converges. Are you using Abel-Dedekind-Dirichlet (summation by parts)? I just figured out how it works, so I had to ask.

 

[daniel_i_l]

I used the Dirichlet Convergence Test:
http://www.mathreference.com/lc-ser,diri.html

 

[Dick]

I used the Dirichlet Convergence Test:
http://www.mathreference.com/lc-ser,diri.html

That's the one. Then, as I said, it looks like it works if you've shown cos(2n) has bounded partial sums.

 

[daniel_i_l]

Well I know that cos(n) has bounded partial sums and cos(2n) = 2(cos(n))^2 - 1 so that means that cos(2n) also has bounded partial sums right?

 

[Dick]

Well I know that cos(n) has bounded partial sums and cos(2n) = 2(cos(n))^2 - 1 so that means that cos(2n) also has bounded partial sums right?

Not at all! cos(n)^2 doesn't have bounded partial sums and neither does 1! How did you show cos(n) had bounded partial sums?? Try and apply the same technique to cos(2n).

 

[HallsofIvy]

Homework Statement

Lets say that I have some sequence $$(a_n)$$ which converges to 0 at infinity and that for all n $$a_{n+1} < a_n$$ but the sequence $$(a_n)$$ diverges.

You mean series rather than that last "sequence", right?