Can we construct a sequence $\{a_{i}\}$, where $0 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 $a_{i}$?
Construction a sequence of real numbers
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calculus
real-analysis
sequences-and-series
convergence
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2Edited 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. – 2012-06-24
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0Oh 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 $a_{i}$? – 2012-06-24
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0About the Edit: try $a_n=B_n-B_{n-1}$. – 2012-06-24
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0?? 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? – 2012-06-24
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0@did: yes, this is working, by taking $a_{i}=\frac{1}{i(i+1)}$, Thanks! – 2012-06-24
1 Answers
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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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0Thank you, this is the good reply! – 2012-06-24
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0What happened to the condition $B_n\to b$? – 2012-06-24
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0@did: That is something one can check independently. – 2012-06-24
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0Michael: 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. – 2012-06-24
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0@did: You are right, the only nonnegative sequence for which this works is constant zero. It should work for some nonpositive sequences though. – 2012-06-24