This is a curiosity question.

Let's consider the following sum: $$S=\sum_{n=0}^{+\infty} \frac{1}{\sqrt{ n !}}$$

The question asked to prove its convergence, which I did using the ratio test. So I tried to find a closed form for the this sum but without luck.

My question

1) Is there a closed form for $S$? (very likely no). 2) If the answer is no can we prove that there is no closed form for $S$ rigorously?


  • 4
    $\begingroup$ To prove or disprove 2) rigorously, there must be a rigorous definition of "closed form". $\endgroup$ – ajotatxe Jun 5 '15 at 17:50
  • $\begingroup$ It is already non-trivial to prove the limit is $e$ without the square-root, if you use the $\lim_{x \to \infty} (1+1/x)^x$ definition of $e$. So I suspect you are right there's probably no closed formula in terms of known constants and functions, or at least, it would be very hard to determine. $\endgroup$ – user2566092 Jun 5 '15 at 17:50
  • $\begingroup$ It seems quite unlikely. See mathforum.org/kb/message.jspa?messageID=5628902 or wilmott.com/messageview.cfm?catid=34&threadid=44616 (which can be found by computing the sum numerically and googling the resulting value). $\endgroup$ – Michael Lugo Jun 5 '15 at 17:53
  • $\begingroup$ @ajotatxe Closed form $=$ combination (sum, product and composition) of some usual functions $F$ applied to some usuel values V. where for example $F=\left\{x\to \sin x,x\to \cos x,x\to e^x\right\}\cup\mathbb{Q}[X]$ and $V=\mathbb{Q}\cup \{e,\pi\}$ $\endgroup$ – Elaqqad Jun 5 '15 at 18:04
  • $\begingroup$ "Closed form number" has (essentially) no non-trivial theory. Unlike "closed form function" which has many interesting non-trivial results. So: even if you prove $\sum x^n/\sqrt{n!}$ is not an elementary function, it does not follow that $\sum 1/\sqrt{n!}$ is "not a closed form number". $\endgroup$ – GEdgar Jun 5 '15 at 18:44

$S(a)=\displaystyle\sum_{n=0}^\infty\frac1{(n!)^a}$ converges for all $a>0$, however, the only known closed forms are $S(1)=e$,

and $S(2)=I_0(2)$, see Bessel function for more information.

| cite | improve this answer | |
  • $\begingroup$ Thanks you for your answer $\endgroup$ – Elaqqad Jun 5 '15 at 18:40

IIRC there's a contour-integral expression for convolution of power series that gives the function $h(x)=f(x)\circ g(x)=\sum_{n=0}^\infty f_ng_nx^n$ in terms of the functions $f(x)=\sum_{n=0}^\infty f_nx^n$ and $g(x)=\sum_{n=0}^\infty g_nx^n$; you can invert that to get an integral equation for the power series $f(x)=\sum_{n=0}^\infty\frac{x^n}{\sqrt{n!}}$ since the convolution $f\circ f$ is equal to $e^x$. I believe it can be proven with some differential Galois theory that $f()$ is non-elementary, though, so this may not be any help in getting special values of the function. For more references, you might try Melzak's Companion to Concrete Mathematics, which uses the convolution formula to get an expression for the aforementioned Bessel functions.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.