Calculate the amount $\sum\limits_{n=0}^{\infty}\frac{1}{(n!)^{2}}$ My attempt:
$$\sum_{k\geq 0}\frac{1}{(k!)^{2}}=\frac{1}{2\pi i}\sum_{k\geq 0}\oint\frac{e^{z}}{k!z^{k+1}}dz=\frac{1}{2\pi i}\oint z^{-1}e^{z}\sum_{k\geq 0}\frac{z^{-k}}{k!}dz=\frac{1}{2\pi i}\oint z^{-1}e^{z}e^{1/z}dz$$
My attempt failed ... Does anyone have an idea how to calculate the amount?
 A: The Bessel function of the first kind and order zero has the series representation
$$J_0(z)=\sum_{n=0}^\infty \frac{(-1)^n}{(n!)^2}\left(\frac z2\right)^{2n}$$
Letting $z=i2$, we see that
$$J_0(i2)=\sum_{n=0}^\infty \frac1{(n!)^2}$$
By definition of the modified Bessel function, $I_0(z)=J_0(iz)$.  Therefore, we have
$$\sum_{n=0}^\infty \frac1{(n!)^2}=I_0(2)$$
A: By the Parseval equality of Fourier series,
$$\begin{align}
\sum_{n=0}^{\infty} \frac{1}{(n!)^2} 
&= \int_0^1 \left|\sum_{n=0}^{\infty}\frac{e^{2\pi int}}{n!} \right |^2\, dt\\\\
&= \int_0^1 |e^{\cos(2\pi t)+i\sin(2\pi t)}|^2\, dt \\\\
&= \int_0^1 e^{2\cos(2\pi t)}\, dt
\end{align}$$
It remains to evaluate the integral.
A: Integrate around the unit circle $|z|=1$.   Then $\frac{1}{2\pi i}\oint z^{-1}e^{z}e^{1/z}dz$ is correct.  But the closed form answer $I_0(2)$ would probably be done by converting to your series...
Or, maybe like this
$$
\frac{1}{2\pi i}\oint_{|z|=1} z^{-1}e^{z}e^{1/z}dz
=\frac{1}{2\pi i}\oint_{|z|=1} \overline{z}e^{z}e^{\overline{z}}dz
\\ =
\frac{1}{2\pi i}\int_0^{2\pi} e^{-i\theta}e^{2\cos\theta}ie^{i\theta}\;d\theta
=\frac{1}{2\pi}\int_0^{2\pi}e^{2\cos\theta}\;d\theta
$$
The value for this, found in tables, is $I_0(2)$.
How is that done?  Probably by expanding that $\cos$ in a series, and finding that the value of this integral is the series $\sum 1/(n!)^2$.
Alternately, you could consider the function
$$
F(z) = \frac{1}{2\pi}\int_0^{2\pi} e^{z\cos\theta} d\theta
$$
and show that it satisfies the differential equation for the Bessel function $I_0(z)$.
