Timeline for Curious double sum identity
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 12, 2022 at 21:59 | comment | added | Jim Bryan | Yes, @mathworker, I think you should write this as an answer and I will accept it. Great job. So easy when you lay it out! I'm a little embarrassed, but I'm still delighted by your answer. | |
| Nov 12, 2022 at 8:55 | comment | added | Daniele Tampieri | @mathworker21, why don't you post your comment as an answer? I'll upvote it. | |
| Nov 12, 2022 at 1:46 | comment | added | Ofir Gorodetsky | @mathworker21 Your $a^{mn}$ in the denominator should be $a^{m+n}$ I think, and then your sum is just $ \sum_{n=1}^{\infty} n/a^n$, squared. This sum is $a/(a-1)^2$ by, say, differentiating $\sum_{n=1}^{\infty} 1/a^n= 1/(a-1)$. | |
| Nov 12, 2022 at 0:36 | comment | added | mathworker21 | Multiply both sides by $2$ and replace one of the (two) double sums with the same double sum except with $n$ and $m$ changed. You then wish to show $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{mn}{a^{mn}} = \frac{a^2}{(a-1)^4}.$$ | |
| Nov 12, 2022 at 0:15 | history | asked | Jim Bryan | CC BY-SA 4.0 |