Can This Seemingly Impossible Mathematical Series Be Solved?

[hagopbul]

i have this problem and i am Trying to solve but it seems impossible so I need help
(even matmatica unable solve it )
seriesSum((2L+1)·(e^((((-((h)^(2)))·L·(L+1))/(2 J K)),L,0,∞)))

or :
seriesSum((2l+1)·(e^((((-((h)^(2)))·l·(l+1))/(2 J K)),l,0,∞)))

SO TRY YOUR LUKE HERE.

 

[HallsofIvy]

i have this problem and i am Trying to solve but it seems impossible so I need help
(even matmatica unable solve it )
seriesSum((2L+1)·(e^((((-((h)^(2)))·L·(L+1))/(2 J K)),L,0,∞)))

or :
seriesSum((2l+1)·(e^((((-((h)^(2)))·l·(l+1))/(2 J K)),l,0,∞)))

SO TRY YOUR LUKE HERE.

I doubt very much that Mathematica can't do it. I suspect you just don't LIKE Mathematica's answer! Since your sum has several powers of L in the numerator, and only constants in the denominator, the individual terms do NOT go to 0 and so the sum does not converge. (I am assuming that you (2 J K) is just the product of three numbers and not some special notation you chose not to define!)

 

[hagopbul]

no you don't understand no body even Mathematica can do it i try
so i need help? even if you don't type 2*j*k
mathematica can't solve it.

 

[Gib Z]

I bet It would be much easier to do if you rewrote it for us without so many Brackets. I am not even going to bother trying to decipher that.

 

[matt grime]

no you don't understand

No, *you* don't understand: since the terms do not tend to zero (according to HallsOfIvy), the sum does not exist. Further, even if it did exist, Mathematica would be able to tell you what it was, but not necessarily in a 'nice' closed formula involving the variables. You have made the common mistake of thinking that some thing is the answer if and only if it is a nice simple expression.