juantheron
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Let $f(m,n) = 3m+n+(m+n)^2.$ Then value of $\displaystyle \sum_{n=0}^{\infty}\;\; \sum_{m=0}^{\infty}2^{-f(m,n)}=$
The discussion centers around the function $f(m,n) = 3m+n+(m+n)^2$ and the evaluation of the double sum $\displaystyle \sum_{n=0}^{\infty}\;\; \sum_{m=0}^{\infty}2^{-f(m,n)}$. Participants explore the behavior of this function and its implications for the convergence of the sum.
Participants express varying approaches to the problem, with no consensus on a definitive method or solution. The discussion remains exploratory with multiple viewpoints on how to tackle the evaluation of the sum.
Participants have not resolved the implications of the function's behavior or the correctness of the numerical approximation. There are also unresolved questions regarding the previous calculations mentioned.
jacks said:Let $f(m,n) = 3m+n+(m+n)^2.$ Then value of $\displaystyle \sum_{n=0}^{\infty}\;\; \sum_{m=0}^{\infty}2^{-f(m,n)}=$
It's from a Putnam. I don't remember the year though.CaptainBlack said:Might I ask where this question comes from?