If $b_j>0$ for every $j$ and if $\sum^{\infty}_{j=1}b_j$ converges then prove that $\sum^{\infty}_{j=1}(b_j)^3$ converges or diverges. If $b_j>0$ for every $j$ and if $\sum^{\infty}_{j=1}b_j$ converges then prove that $\sum^{\infty}_{j=1}(b_j)^3$ converges or diverges.
I believe it to be true if the positivity hypothesis holds, but I am not sure otherwise...
What if $b_j=\frac{-1^j}{\sqrt{j}}$ then it converges by the zero test but the square of that doesn't converge via the p-test... I am confused how without the positivity hypothesis, we can make the claim that $b_j^3$ converges. If we could always say $b_j^2<b_j$ I would buy it but I don't know if that is the case...
Without the positivity hypothesis I would conjecture that there is convergence as well. I am having trouble coming up with a good proof but my thoughts are as follows. Since 3 is an odd power (like 1), the signs of $b_j^3$ will alternate (if the did for the $b_j$) in the same way as the $b_j$. Furthermore, as $j$ gets large, the $b_j\rightarrow 0$ and cubing a number less than 1 makes the number smaller. So the values should just go to zero faster and respecting the sign changes. Thus, it would converge in theory

If $\sum b_j$ converges then $b_j\rightarrow 0$.
For $j$ large enough you have $0<b_j<1 \Rightarrow 0<b_j^2<b_j$
By multiplying with $b_j$ again you get $0<b_j^3<b_j^2\Rightarrow 0<b_j^3<b_j$
Then $\sum b_j^3$ converges by comparison test.
