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**Plato** Here is the problem: $\displaystyle a_n > a_{n + 1} > 0\,,\,\sum\limits_{n = 1}^\infty {a_n } \,\& \,\sum\limits_{n = 0}^\infty {2^n a_{2^n } } $.

Using that $\displaystyle a_n$ is decreasing, then following shows it in one direction.

$\displaystyle \begin{gathered} a_1 + \underbrace {a_2 + a_3 }_{} + \underbrace {a_4 + a_5 + a_6 + a_7 }_{} + \underbrace {a_8 + a_9 + a_{10} + a_{11} + a_{12} + a_{13} + a_{14} + a_{15} }_{} + \cdots \hfill \\

\leqslant a_1 + 2a_2 + 4a_4 + 8a_8 \cdots \hfill \\

\end{gathered} $

Now note that:

$\displaystyle \begin{gathered} a_2 + \underbrace {a_3 + a_4 }_{} + \underbrace {a_5 + a_6 + a_7 + a_8 }_{} + \underbrace {a_9 + a_{10} + a_{11} + a_{12} + a_{13} + a_{14} + a_{15} + a_{16} }_{} + \cdots \hfill \\ \geqslant a_2 + 2a_4 + 4a_8 + 8a_{16}\cdots \hfill \\ \end{gathered} $

Also note that if $\displaystyle \sum\limits_{n = 1}^\infty {a_n } $ converges that $\displaystyle

\sum\limits_{n = 2}^\infty {2a_n } $ converges.

Can you finish?