Let $1>f(k+1,n)>f(k,n)>0$ and $0 And $f(k,\cdot)\to1, f(\cdot,n)\to0$. Is there an $f$ for every $0
Diagonal triple limit
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real-analysis
sequences-and-series
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0Off-topic: You may use `$\to$` for getting short right arrows instead of `->`. – 2012-02-03
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0Op has now edited his question, in case it goes unnoticed and this comment is noticed! =) – 2012-02-03
1 Answers
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$f(k,n)=c^{n/k}$ works for the original question.
For the $f(k,\cdot)\to\infty$ version mentioned in a (now deleted) comment, $f(k,n)=2^{k-n}c^{n/k}$ works.