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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?

Fk-91
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  • The result can be written in terms of the modified Bessel function as $I_0(2)$. – Mark Viola Dec 19 '21 at 16:49
  • I believe you need division by $(k!)^3$ in your integral – FShrike Dec 19 '21 at 16:52
  • Obviously, $2 < s < e $ and since the terms decrease so fast, it's a little more than 2.25. Wolfram Alpha calculates it as the value at 2 of a modified Bessel function: $I_0(2)$. See https://functions.wolfram.com/Bessel-TypeFunctions/BesselI/06/ShowAll.html – NickD Dec 19 '21 at 16:53
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    Your attempt calls out a contour integral without indicating what contour. There are several contours where that integral is surprisingly easy. There are also contours where the integral gives the value you want. Your reader is not responsible for guessing which you mean. – Eric Towers Dec 19 '21 at 17:25
  • Please include the problem statement in the body of your Question. It is okay to put it in the title, but the body of your Question permits more space (and room for context, such as why the problem is important or interesting). – hardmath Dec 22 '21 at 23:45

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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

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

Mark Viola
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  • I've made my point clear that it would be nice if properties of Bessel functions could be used to obtain some information about the original series. And yes, I say that you have proved nothing in this post: you defined $J_0(z)$ as a power series, then you set $z = 2i$ and recover the original series. It is equally useless as naming $\sum_{n = 0}^\infty \frac 1{n!} = e$ without giving any of the useful properties of $e$, but it would be a different story if you could prove e.g. $\lim_{n \rightarrow \infty}(1 + \frac1n)^n = e$ at the same time. – WhatsUp Dec 19 '21 at 19:52
  • Not sure who deleted my (and the the other person's) previous comment. I don't think my comment contained anything rude or unpleasant - I'm in a mood of discussing mathematics. Although I thought that the other person's comment was not so friendly. – WhatsUp Dec 19 '21 at 19:55
  • @WhatsUp Another false claim of yours needs refutation. I never defined $J_0(z)$ as a power series. I clearly stated that $J_0(z)$ has a series representation as given. The Bessel functions can be defined as solutions to Bessel's differential equation. And one can derive the series representation using a standard methodology (Frobenius) Once equipped with that representation, one can evaluate the Bessel function at any point in its domain using the series. That is what I did here. What on earth is preventing you from understanding something so elementary? – Mark Viola Dec 19 '21 at 20:41
  • @WhatsUp Furthermore, in your comment you imply that one could prove that $\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$. In order to do so, one would need a definition of $e$. There are several characterizations of the function $\exp(x)$ from which one can define $e$ as $e=\exp(1)$. One such characterization is the solution to the ODE $y'=y$, $y(0)=1$. Another is the Taylor series of $\exp(x)$. A third is the limit $\lim_{n\to\infty}\left(1+\frac xn\right)^n$. And one can show these are equivalent. – Mark Viola Dec 20 '21 at 02:26
  • So, this is, in kind, no different from defining the Bessel function $J_0(x)$ as the solution of Bessel's differential equation, OR a power series, OR another representation (e.g., an integral representation). All of these can be shown equivalent. And once equipped with one, say the Taylor series representation I used, one can evaluate the funciton at any point in its domain. Here, we began with the Taylor series and evaluated it at $i2$. This yields $$J_0(i2)=\sum_{n\ge0}\frac1{(n!)^2}$$which is the series of interest. You should now see that clearly your assertions are flawed. – Mark Viola Dec 20 '21 at 02:31
  • @WhatsUp Have a Look at This Post and the Accepted Answer Therein. Do you see anything that looks familiar? They begin with an integral representation and arrive at a series representation of ... WAIT FOR IT ... The Modified Bessel Function of the First Kind and Order Zero. And the microphone drops. ;-)) – Mark Viola Dec 21 '21 at 04:55
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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.

Mark Viola
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Onur Oktay
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  • I really like your idea of using Parseval. To conclude, note that the last integral is connected to the Bessel function. See the Wikipedia article about it – Stefan Lafon Dec 19 '21 at 18:16
  • The last integral is an integral representation of the modified Bessel function of the first kind and order $0$, evaluated at $2$. That is, the result is $I_0(2)$. – Mark Viola Dec 19 '21 at 19:25
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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 $I_0(z)$.

GEdgar
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