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double sum

jacks

Well-known member
Apr 5, 2012
226
Let $f(m,n) = 3m+n+(m+n)^2.$ Then value of $\displaystyle \sum_{n=0}^{\infty}\;\; \sum_{m=0}^{\infty}2^{-f(m,n)}=$
 

Krizalid

Active member
Feb 9, 2012
118
Take a look at $f(n,m).$
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Let $f(m,n) = 3m+n+(m+n)^2.$ Then value of $\displaystyle \sum_{n=0}^{\infty}\;\; \sum_{m=0}^{\infty}2^{-f(m,n)}=$
Might I ask where this question comes from?

The sum converges very quickly and can be evaluated numerically with 4 terms of each summation (it is \(\approx 1.33333\), which is very suggestive ... ) to good accuracy.

CB

(why the previous calc got the wrong answer I still have no idea)
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
I hope Krizalid will not object if I add a bit to his very helpful suggestion. In problems like this, my advice is always to start by looking at what happens for small values of the variables. In this case, if you make a table of the values of $f(m,n)$ for small values of $m$ and $n$, it looks like this:

$$\begin{array}{cc}&\;\;\;\;n \\ \rlap{m} & \begin{array}{c|cccc} &0&1&2&3 \\ \hline 0&0&2&6&12 \\ 1&4&8&14&. \\ 2&10&16&.&. \\ 3&18&.&.&. \end{array} \end{array}$$

Doesn't that suggest something very interesting about the range of the function $f(m,n)$?
 

Krizalid

Active member
Feb 9, 2012
118