$\sum_{k=1}^\infty a_k$ converges if and only if $\sum_{k=1}^\infty 2^k*a_{2^k}$ Suppose $a_k$ $\ge$ $0$, for $k$ $\ge$ $1$, and $a^{k's}$ are decreasing in $k$. Show that $\sum_{k=1}^\infty a_k$ converges if and only if $\sum_{k=1}^\infty 2^k*a_{2^k}$ converges. I was asked to proof this but I am stuck. 
The ratio test didn't help me. So I thought about the comparison test. However, using the comparison test at most I can proof the statement one way. Does someone have an idea how to prove this?

This is Cauchy condensation; see that link for a derivation, which you'll understand if you know the most popular proof the harmonic series diverges. But in short, $$a_1+\sum_{j=2}^n2^{j-1}a_{2^j}=a_1+\sum_{k=2}^{2^n-1}a_{2^{\lceil\log_2k\rceil}}\le a_1+\sum_{k=2}^{2^n-1}a_k\\=\sum_{k=1}^{2^n-1}a_k\\\le\sum_{k=1}^{2^n-1}a_{2^{\lfloor\log_2k\rfloor}}\le\sum_{j=1}^n2^{j-1}a_{2^{j-1}}.$$
