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What is the closed form for $\sum_{n=0}^{\infty}\frac{1}{(n!)^2}$? And is there a closed form for $\displaystyle \sum_{n=0}^{\infty}\frac{1}{(n!)^k}$?

Edit: If there are no closed forms for the two series, how should I convert them into integrals? Now I know that $\displaystyle I_0(2)=\sum_{n=0}^{\infty}\frac{1}{(n!)^2} = \frac{1}{\pi}\int_{0}^{\pi}e^{2\cos\theta}d\theta$, but is there an integral representation for $\displaystyle \sum_{n=0}^{\infty}\frac{1}{(n!)^k}$?

Larry
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    $I_0(\sqrt2)$ where $I_0$ is a modified Bessel function. – Angina Seng Sep 02 '18 at 13:45
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    I think it is $I_0(2)$ instead of $I_0(\sqrt{2})$ right? – Larry Sep 02 '18 at 15:13
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    No \displaystyle in titles please. – Did Sep 03 '18 at 12:07
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    An integral representation for $ \sum_{n=0}^{\infty}\frac{1}{(n!)^k}$ seems to be an interesting question. Voted to reopen. – GEdgar Sep 03 '18 at 12:26
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    It would help to add some context to this Question (as with others posted by the OP). What motivates the interest in series of this form? Is there a relationship underlying other recent posts by this OP about series and closed forms? Such context may help motivate Readers to invest time in researching problems. – hardmath Sep 16 '18 at 15:04
  • The function $u(x):=\sum_{n=0}^\infty x^n/(n!)^3$ solves the linear ODE $(x(xu')')'=u$. Curiously, solutions of $(x(xu')')'=0$ would be easier to find: they are $P(\log x)$ with $P$ polynomial of degree less than $3$. Analogously for any $k$. – Pietro Majer Aug 08 '19 at 15:18
  • By definition, ${0 F{k - 1}}(; \mathbf 1; 1) = \sum_{n \geq 0} 1/(n!)^k$. – Maxim Aug 08 '19 at 23:37

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