MHB Sum of a discrete finite sequence

AI Thread Summary
The discussion focuses on finding a tight upper bound for the sum of the sequence {bi} defined as bi = ai * (1/2^i), given the sum of the sequence {ai} is known (x). It is noted that if the terms of {ai} are non-negative, the sum of {bi} can range between x/2^k and x/2, depending on the distribution of the ai values. The extreme cases illustrate that without additional information about the ai values, a tighter bound cannot be established. The attempt to use the convexity of 1/2^i did not yield conclusive results. Overall, the conclusion is that more information about the sequence {ai} is necessary for a precise calculation of the sum of {bi}.
bincy
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Hii everyone,

I have a sequence {ai,1<= i <=k} where i know the sum of this sequence(say x).
I want to know the sum of another sequence {bi, 1<=i <=k}(at least a tight upper bound) where bi=ai*(1/2^i).

Or in other words, if you know the sum of the ratio sequence and sum of 1 sequence, how to find out the sum of the other sequence(can we)?

I tried using the convexity of 1/2^i, but couldn't get anything.

regards,
Bincy.
 
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bincybn said:
Hii everyone,

I have a sequence {ai,1<= i <=k} where i know the sum of this sequence(say x).
I want to know the sum of another sequence {bi, 1<=i <=k}(at least a tight upper bound) where bi=ai*(1/2^i).

Or in other words, if you know the sum of the ratio sequence and sum of 1 sequence, how to find out the sum of the other sequence(can we)?

I tried using the convexity of 1/2^i, but couldn't get anything.
In general, there is very little that you can say about $\sum_{i=1}^ka_i/2^i$.

I assume that the terms in the sequence $\{a_i\}$ are non-negative (if not, then there is even less that you can say about the sum of the series). If you think about the possible values of the $a_i$ (subject to the condition that their sum is $x$), then at one extreme you could have $a_1=x$ and $a_i=0$ for $2\leqslant i\leqslant k$. At the other extreme you could have $a_k=x$ and $a_i=0$ for $1\leqslant i\leqslant k-1$. In the first case, $\sum_{i=1}^ka_i/2^i = x/2$. In the second case, $\sum_{i=1}^ka_i/2^i = x/2^k$. So (unless you have further in formation about the $a_i$), all you can say about the sum of the $b_i$ is that it lies between $x/2^k$ and $x/2$. Presumably that does no qualify as a tight bound!
 
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