$(a_k)_{k\geq 1} $ is a monotonically decreasing sequence with numbers $ \geq 0$.
Prove: The series $\sum_{k=1}^{\infty} a_k $ convergent $\Leftrightarrow$ $\sum_{k=1}^{\infty} 2^k*a_{2^k}$ convergent.
My toughts were that if the series on the left side is monotonically decreasing and the series is infinite, from a certain point a partial sum would not grow anymore so the row is limited. But i cant express myself in a mathematical way...
Sketch (for one direction):
Write (since the terms of the sum are all positive, the sum is absolutely convergent, so we can rearrange the terms): $$ \sum_{k=1}^\infty a_k=\sum_{j=0}^\infty\sum_{k=2^j}^{2^{j+1}-1}a_k $$
Since the sum is decreasing, we know that if $k\leq2^{j+1}$, then $a_k>a_{2^{j+1}}$. Therefore, $$ \sum_{j=0}^\infty\sum_{k=2^j}^{2^{j+1}-1}a_k\geq \sum_{j=0}^\infty\sum_{k=2^j}^{2^{j+1}-1}a_{2^{j+1}}=\sum_{j=0}^\infty 2^ja_{2^{j+1}}=\sum_{j=1}^\infty 2^{j-1}a_{2^j}. $$
Now, just multiply the sum by $2$ to get your desired sum.
For the other direction, try a similar trick (bound the sums above by $2^ja_{2^j}$).
