Evaluating $\sum_{n=0}^{\infty}\frac{J_n(n)}{n!}$

Does anyone have advice on how to go about finding (if it exists) a closed form for

$$\sum_{n=0}^{\infty}\frac{J_n(n)}{n!}$$

Where $J_n$ represents the Bessel function of the first kind; numerically it appears to converge to $\approx 1.68226...$

I tried playing around with the recurrence relations for the Bessel functions in an attempt to find an exponential generating function for $J_n(n)$, but this lead to nonsensical results. Any other ideas?

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We have that $$J_n(x) = \dfrac1{2 \pi} \int_{-\pi}^{\pi} e^{-i(n t-x \sin(t))} dt$$ Hence, $$J_n(n) = \dfrac1{2 \pi} \int_{-\pi}^{\pi} e^{-in(t-\sin(t))} dt = \dfrac1{2 \pi} \int_{-\pi}^{\pi} (a(t))^{n} dt$$ where $e^{-i(t-\sin(t))} = a(t) = \cos(t-\sin(t))-i\sin(t-\sin(t))$. We then have $$\sum_{n=0}^{\infty} \dfrac{J_n(n)}{n!} = \dfrac1{2\pi} \int_{-\pi}^{\pi} \sum_{n=0}^{\infty} \dfrac{(a(t))^{n}}{n!} dt = \dfrac1{2 \pi} \int_{-\pi}^{\pi} \exp(a(t)) dt\\ = \dfrac1{2 \pi} \int_{-\pi}^{\pi} \exp(\cos(t-\sin(t))) \cdot \cos(\sin(t-\sin(t))) dt$$which approximately evaluates to $1.68226$.

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Thanks! It's nice how that works out. – Bitrex Jan 13 '13 at 5:27
@Bitrex There might be some neat "closed" form way to evaluate the integral as well. Nothing comes to my mind now though. – user17762 Jan 13 '13 at 5:28

In series form:

$\sum\limits_{n=0}^{\infty}\dfrac{J_n(n)}{n!}=\sum\limits_{n=0}^{\infty}\sum\limits_{k=0}^\infty\dfrac{(-1)^kn^{n+2k}}{n!k!(n+k)!2^{n+2k}}$

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Unfortunately, I don't see how this leads to the evaluation of a closed form. – Bitrex May 16 '15 at 16:16