$b_j>0$ 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??


You can't say.

$\sum 1 =\infty$ and $\sum 2^j = \infty$ but $\sum 2^{-j} \times 1=1$ and $\sum 2^{-j} \times 2^j=\infty$

  • $\begingroup$ 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..... $\endgroup$ – Fernando Martinez Sep 25 '15 at 23:04
  • $\begingroup$ @FernandoMartinez I'm sorry I don't understand your statement. $\sum 2^{-j}=1$. $\endgroup$ – user223391 Sep 25 '15 at 23:05
  • $\begingroup$ Oh yes you are right.... $\endgroup$ – Fernando Martinez Sep 25 '15 at 23:07
  • $\begingroup$ Yes it is a geometric sum $\endgroup$ – Fernando Martinez Sep 25 '15 at 23:18
  • $\begingroup$ So you have a divergent sum times a convergent sum and then you get a divergence sum because $1*\infty$ is infity?? $\endgroup$ – Fernando Martinez Sep 25 '15 at 23:19

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