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?

Solution 1:

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

$$\sum_{n=0}^\infty \frac1{(n!)^2}=I_0(2)$$

Solution 2:

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.

Solution 3:

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 $I_0(z)$.