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Is there some manipulation $g$ that can be applied to the sum $$I(n) = \sum_{J=n+3}^{\infty} \frac{f(n,J)}{(J-n-2)!}$$

Such that we are left with $$g(I(n)) = \sum_{J=n+3}^{\infty} \frac{f(n,J)}{(J-n)!}$$

edit: $f(n,J) = \frac{(-1)^{J+n} \zeta(J) (J-1)! 2^{n}}{2^{J} (2^{n} - 2)}$

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    What do we know about $f(n,J)$? Any link between $f(n,J+2)$ and $f(n,J)$? – 2017-02-14
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    $f(n,J) = \frac{(-1)^{J+n} \zeta (J) (J-1)! 2^{n}}{2^{J} (2^{n} -2)}$ – 2017-02-14
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    @Riakm That should've already been in the question. – 2017-02-14
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    By the change of index $j=J-2$ one gets $$\sum_{J=n+3}^{\infty} \frac{f(n,J)}{(J-n-2)!}=\sum_{j=n+1}^{\infty} \frac{f(n,j+2)}{(j-n)!} $$ I don't know if this can help. – 2017-02-14

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Let \begin{eqnarray*} K(n,x)= \sum_{J=n+3}^{\infty} \frac{f(n,J)}{(J-n)!} x^{J-n} \end{eqnarray*} Now differentiate twice with respect to $x$ and we have \begin{eqnarray*} K''(n,x)= \sum_{J=n+3}^{\infty} \frac{f(n,J)}{(J-n-2)!} x^{J-n-2} \end{eqnarray*} Now set $x=1$.

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    The main problem I see with this approach is that I do not know the closed form of $K''(n,x)$ only $K''(n,1)$, and thus I cannot integrate to find $K(n,1)$. – 2017-02-14
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    Is $\zeta(J)$ the Riemann zeta function ? by any chance ? – 2017-02-14
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    Yes $\zeta(J)$ is the Riemann Zeta function. – 2017-02-14