# Divergence or convergence $\sum 2^{-j}b_j$

If

$b_j>0$ and $\sum b_j$ diverges then

does the sum of

$\sum 2^{-j}b_j$

converge or diverge.

If $b_j$ diverges then limit is something that is not zero.

I would say the series

as $\sum \frac{b_j}{2^j}$ is like the harmonic series it diverges??

$\sum 1 =\infty$ and $\sum 2^j = \infty$ but $\sum 2^{-j} \times 1=1$ and $\sum 2^{-j} \times 2^j=\infty$
• Hmm but I know $\sum 2^-j$ is infinity and $b_j$ is divergent so when you combine them both you get infity.... or not ..... maybe..... – Fernando Martinez Sep 25 '15 at 23:04
• @FernandoMartinez I'm sorry I don't understand your statement. $\sum 2^{-j}=1$. – user223391 Sep 25 '15 at 23:05
• So you have a divergent sum times a convergent sum and then you get a divergence sum because $1*\infty$ is infity?? – Fernando Martinez Sep 25 '15 at 23:19