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Can we construct a sequence $\{a_{i}\}$, where $0<a_{i}\leq 1$ for all $i$, such that $\sum_{i=1}^{n}a_{i}=\frac{B_{n}}{2^{n}}$ such that $B_{n}\to b$ as $n\to \infty$ for some $1\leq b <\infty$?

Edit: What about $\sum_{i=1}^{n}a_{i}=\frac{n}{n+1}$, or $\sum_{i=1}^{n}a_{i}=B_{n}$ with $B_{n}\to b$, could we find such $a_{i}$?

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Edited twice while I was trying to read it = most annoying. If $B_n\to b$ then the right side goes to zero, but the left side doesn't. –  Gerry Myerson Jun 24 '12 at 8:09
    
Oh I'm so sorry, my bad! So what about $\sum_{i=1}^{n}a_{i}=\frac{n}{n+1}$, or $\sum_{i=1}^{n}a_{i}=B_{n}$ with $B_{n}\to b$, could we find such $a_{i}$? –  Ben Jun 24 '12 at 8:19
    
About the Edit: try $a_n=B_n-B_{n-1}$. –  Did Jun 24 '12 at 8:26
    
?? What is your question, in the end? Given $(B_n)$, find $(a_n)$, or, given $(a_n)$, find $(B_n)$, or find $(B_n)$ with $B_n\to b$ for some given $b$, or what? –  Did Jun 24 '12 at 8:30
    
@did: yes, this is working, by taking $a_{i}=\frac{1}{i(i+1)}$, Thanks! –  Ben Jun 24 '12 at 8:30
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Whether $(B_n)$ converges or not does not really matter. That such a sequence exists is equivalent to $$0<\underbrace{\frac{B_{n+1}}{2^{n+1}}-\frac{B_n}{2^n}}_{\text{ this would be }a_{n+1}}\leq 1,$$ for $n\geq 1$, since this uniquely determines the sequence $(a_n)$ as above with $a_1=B_1/2$.

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Thank you, this is the good reply! –  Ben Jun 24 '12 at 8:36
    
What happened to the condition $B_n\to b$? –  Did Jun 24 '12 at 8:54
    
@did: That is something one can check independently. –  Michael Greinecker Jun 24 '12 at 9:09
    
Michael: My question was directed at Ben more than at you, nevertheless: if $(B_n/2^n)_n$ is increasing and $B_1=2a_1\gt0$, I wonder how $(B_n)$ can possibly converge to a finite limit. –  Did Jun 24 '12 at 9:13
    
@did: You are right, the only nonnegative sequence for which this works is constant zero. It should work for some nonpositive sequences though. –  Michael Greinecker Jun 24 '12 at 9:38
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