Interesting Summation?

I came across a rather interesting identity while messing around with one of my favorite functions (which will be left unidentified, for now). The identity, in text, is sum{sum{ (-2)^n*x^(2n+2k+1)/[k! (2n+1)!!] }} = sum{x^(2m+1)/[(2m+1) m!]} the summations are taken wrt the appropriate variables n, k, and m,... show more I came across a rather interesting identity while messing around with one of my favorite functions (which will be left unidentified, for now). The identity, in text, is

sum{sum{ (-2)^n*x^(2n+2k+1)/[k! (2n+1)!!] }}
= sum{x^(2m+1)/[(2m+1) m!]}
the summations are taken wrt the appropriate variables n, k, and m, and all run from 0 to infinity.

image version:
http://www.ilstu.edu/~bmreini/sumid.bmp

I'd like to see someone give a direct proof of the identity. I want to know if my round-about discovery of this is actually better than trying to show it directly. Bonus points if anyone can deduce how I came up with the thing in the first place.

(Oh, and that is a double factorial in the LHS; that means the product of the odd integers up to 2n+1 in this case, NOT the factorial of the factorial.)

I'll try to put up some helpful hints up here every day or so...
Update: Very nice A Cave! In the spirit of keeping the question alive, I'm still interested in seeing a proof of the identity by matching coefficients...in essence, prove that the two expressions are equivalent as power series. Also, how might I have come up with the double summation LHS?...you've already... show more Very nice A Cave!

In the spirit of keeping the question alive, I'm still interested in seeing a proof of the identity by matching coefficients...in essence, prove that the two expressions are equivalent as power series.

Also, how might I have come up with the double summation LHS?...you've already noted that LHS = exp(x^2)*stuff, so perhaps this won't be too hard.
Update 2: Methinks people have forgotten this question...

Anyway, here's a hint for the coefficient matching: make a change of variables in the double summation. In particular, notice that if we sum over (n+k), the powers of x would match up nicely...in fact, more things than that start to match...
Update 3: Hmm, more stars...that means people are still around. The next hint (with 3 days to answer): Using the last hint, we find that it is equivalent to show sum{ (-4)^n*n! / [(m-n)! (2n+1)!] } = 1/[(2m+1) m!] where the sum is wrt n from 0 to m. http://www.ilstu.edu/~bmreini/sumid2.bmp This is where I'm... show more Hmm, more stars...that means people are still around.

The next hint (with 3 days to answer):
Using the last hint, we find that it is equivalent to show
sum{ (-4)^n*n! / [(m-n)! (2n+1)!] } = 1/[(2m+1) m!]
where the sum is wrt n from 0 to m.
http://www.ilstu.edu/~bmreini/sumid2.bmp

This is where I'm out...I have some movement forward from here, but nothing that seems relevant.
1 answer 1