2
$\begingroup$

I would like to calculate, or find a sharp upper bound for the following sum: $ \sum_{i=0}^m \frac{\sqrt{a_{i+1}}}{2^i}, $ if $m\leq b=\sum_{j=1}^{m+1}a_j$.

  • 0
    @bgins: yes, it is the same $m$. All I know is that $a_i$ are non- negative numbers. We can wven assume that they are naturals with zero.2012-04-07

1 Answers 1

1

If $a_1=b, a_j=0$ for $j \gt 1$, then the sum is $ \frac{\sqrt b}2$. Dropping $a_1$ and increasing the others will only reduce the sum. If some of the $a$'s can be negative, I don't think there is a limit: Let $a_1$ increase positive and $a_2$ increase negative keeping $b$ the same.