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While solving a few sums, I came across the following double sum $$\sum_{k=0}^{n} (-1)^k\frac{(n+k)!}{2^kk!(n-k)!}x^{n-k}\sum_{j=0}^{n+k} \frac{x^j}{j!}$$

which is expected to evaluate to $$\sum_{k=0}^n (-1)^{n-k} \frac{(2(n-k)-1)!!}{(2k)!!}x^{2k}$$ where $n!!$ is the double factorial.

Question

I have no idea how to prove this, so any hints or answers would be really appreciated!

see also related post.

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  • $\begingroup$ This seems to be true only if $n$ is even or exponent of $x$ is odd. Consider e.g. the coefficients of $x^0$ when $n=1$. $\endgroup$ Commented Sep 4 at 16:52
  • $\begingroup$ your last expression, with the Struve function, is not correct, see mathoverflow.net/a/500875/11260 for the correct formula $\endgroup$ Commented Sep 26 at 11:56
  • $\begingroup$ Thanks Carlo, I corrected it. $\endgroup$ Commented Sep 26 at 11:57
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    $\begingroup$ We're now up to 18 edits. Each edit pushes the question to the top of the queue, and thus should only be done for important reasons. I think it is time to stop editing. $\endgroup$ Commented Nov 4 at 18:12
  • $\begingroup$ @AndyPutman Sorry, last time. $\endgroup$ Commented Nov 5 at 20:19

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Make the change of summation variables $j=m-l$, $k=n-l$. After simplification, your sum becomes $$(-1)^n\frac{(2n)!}{2^n n!}\sum_{m=0}^{2n}\frac{x^m}{m!}\sum_{l=0}^{\min(m,n)}\frac{(-m)_l(-n)_l}{l!(-2n)_l}\,2^l. $$ The inner sum is a special case of Gauss's hypergeometric series. By identity 7.3.8.2 in Prudnikov's Integrals and Series, Volume 3, it equals $0$ if $m$ is odd and $$\frac{(1/2)_{m/2}}{(-n+1/2)_{m/2}}=(-1)^{m/2}\frac{m!\,n!(2n-m)!}{(m/2)!(2n)!(n-m/2)!} $$ else. Writing $m=2k$, it follows that your sum equals $$\sum_{k=0}^n\frac{(-1)^{n+k}}{2^n}\frac{(2n-2k)!}{k!(n-k)!}\,x^{2k}.$$ This differs from your right-hand side by the factor $(-1)^n$, which I guess is a typo.

Is this related to the question Reducing a triple combinatorial sum to a single sum? Curiously, I answered it yesterday using exactly the same hypergeometric identity.

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  • $\begingroup$ Wow ! No it's not related to it, it's a completly independant question (I'm not aware of that one you hint); look, i'll rewrite the original first sum because i invert it myself; can you look again. Thanks. $\endgroup$ Commented Sep 5 at 11:33
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    $\begingroup$ Now you have changed the notation in your question so that my answer no longer makes sense. The previous version was fine except that you had forgotten a factor $(-1)^n$. $\endgroup$ Commented Sep 5 at 11:46
  • $\begingroup$ it's not changed at all, just this is the original sum I had, and the one written previously is the inverted one, so I understand from your response that I have to invert it, no problem i'll do it in the question. Thank you! $\endgroup$ Commented Sep 5 at 11:51
  • $\begingroup$ I am just asking if we can begin from the original sum without inverting, or we must invert first and then do your calculation, just asking about the most natural way to do it. $\endgroup$ Commented Sep 5 at 11:56
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    $\begingroup$ You claim that $$\sum_{j=0}^{2n} \frac{x^j}{j!}\sum_{k=\max(0,j-n)}^{n} (-1)^k\frac{(n+k)!}{2^kk!(n-k)!}x^{n-k}=\sum_{k=0}^n (-1)^k \frac{(2(n-k)-1)!!}{(2k)!!}x^{2k}.$$ The correct identity is $$\sum_{j=0}^{2n} \frac{x^j}{j!}\sum_{k=\max(0,j-n)}^{n} (-1)^k\frac{(n+k)!}{2^kk!(n-k)!}x^{n-k}=(-1)^n\sum_{k=0}^n (-1)^{k} \frac{(2(n-k)-1)!!}{(2k)!!}x^{2k}.$$ $\endgroup$ Commented Sep 5 at 12:17

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