A question about convergence test in $2^n$ form For a decreasing and non-negative sequence {$a_n$}, we have: 

if $\displaystyle \sum_{k=1}^{\infty}2^k \cdot a_{2^k}$ converges, then $\displaystyle \sum_{k=1}^{\infty}a_{k}$ also converges.

Is the converse of this statement also write? That is:

If $\displaystyle \sum_{k=1}^{\infty}a_{k}$ converges, does $\displaystyle \sum_{k=1}^{\infty}2^k \cdot a_{2^k}$ also converge? And why?

Thank you in advance.
$2a_2&gt;a_2+a_3$

Simularly $4a_4&gt;a_4+a_5+a_6+a_7$ 

$8a_8&gt;a_8+a_9+a_{10}.......a_{15}$

Hence S= $\sum_{k=1}^{\infty}2^k \cdot a_{2^k} &gt; a_2+a_3 +a_3.......$ ,  we know S and $a_1$ are finite hence proved.



$2a_2=2a_2$

$2a_3+2a_4&gt;4a_4$

$2a_5+2a_6+2a_7+2a_8&gt;8a_8$

Therefore 
$(2 \sum_{k=1}^{\infty} a_k ) - 2a_1 &gt; 
\sum_{k=1}^{\infty}2^k \cdot a_{2^k} $

So the converse is also true.
$2a_2&gt;a_2+a_3$

Simularly $4a_4&gt;a_4+a_5+a_6+a_7$ 

$8a_8&gt;a_8+a_9+a_{10}.......a_{15}$

Hence S= $\sum_{k=1}^{\infty}2^k \cdot a_{2^k} &gt; a_2+a_3 +a_3.......$ ,  we know S and $a_1$ are finite hence proved.



$2a_2=2a_2$

$2a_3+2a_4&gt;4a_4$

$2a_5+2a_6+2a_7+2a_8&gt;8a_8$

Therefore 
$(2 \sum_{k=1}^{\infty} a_k ) - 2a_1 &gt; 
\sum_{k=1}^{\infty}2^k \cdot a_{2^k} $

So the converse is also true.