R rupa New member Joined Feb 4, 2015 Messages 1 Feb 4, 2015 #1 \(\displaystyle \displaystyle \sum_{m,n \geq 0} \dfrac{mn}{(m+n)!} \)
Steven G Elite Member Joined Dec 30, 2014 Messages 14,605 Feb 4, 2015 #2 rupa said: \(\displaystyle \displaystyle \sum_{m,n \geq 0} \dfrac{mn}{(m+n)!} \) Click to expand... I take it that m and n are integers?? What are we summing over? Or are we summing over all m,n combinations where m and n are positive integers?
rupa said: \(\displaystyle \displaystyle \sum_{m,n \geq 0} \dfrac{mn}{(m+n)!} \) Click to expand... I take it that m and n are integers?? What are we summing over? Or are we summing over all m,n combinations where m and n are positive integers?
stapel Elite Member Joined Feb 4, 2004 Messages 16,543 Feb 5, 2015 #3 rupa said: How to do it? \(\displaystyle \displaystyle \sum_{m,n \geq 0} \dfrac{mn}{(m+n)!} \) Click to expand... What is the "it" what you're needing to "do"? What were the instructions? What have you tried? Where are you stuck? Please be complete. Thank you!
rupa said: How to do it? \(\displaystyle \displaystyle \sum_{m,n \geq 0} \dfrac{mn}{(m+n)!} \) Click to expand... What is the "it" what you're needing to "do"? What were the instructions? What have you tried? Where are you stuck? Please be complete. Thank you!