Find zero of function $\sum_{n=0}^{\infty}\frac{cos(x(n+1))}{n!}$ I'm interested in finding the smallest positive zero of function
$$\sum_{n=0}^{\infty}\frac{\cos(x(n+1))}{n!},\qquad  x\in\mathbb R$$
It is approximately equal to $0.832$. I've calculated Taylor series expansion of this sum, which is:
$$\sum_{k=0}^{\infty}e \cdot\frac{(-1)^k \ B_{2k}\ x^{2k}}{(2k)!}$$
Where $B_n$is n-th Bell number.
But I don't know, whether it helps.
Thanks for all the help in solving this problem.
 A: The more exact value is $\;0.83171119357973597757600960396587803808517294078679544552179166\cdots
$ which is equal to $\;\dfrac{\pi}2-D\;$ with $D$ the "Dottie Number".
The Dottie number is obtained by iteration of the $\cos$ function on a calculator (in 'radian' mode) as detailed in Mathworld.
A: Your sum is equal to
$$\Re\bigg(\sum_{n=0}^\infty \frac{e^{ix(n+1)}}{n!}\bigg)$$
or
$$\Re\exp\big(ix+e^{ix}\big)=\Re\exp\big(\cos x+i\sin x+ix\big)$$
Using Euler’s formula, this is equal to
$$e^{\cos x}\cos(x+\sin x)$$
To find when this is equal to zero, you must compute the zeroes of the function
$$\cos(x+\sin x)$$
or the values of $x$ for which
$$x+\sin x = \pi(n+1/2)$$
for some $n\in\mathbb Z$. I do not suspect an elementary solution will exist, but you can use this explicit form to calculate some pretty good approximate solutions.
The smallest of the zeroes will occur when
$$x+\sin x = \pi/2$$
which is at approximately $x\approx 0.832$. See @Raymond’s answer for a more accurate approximation and a representation in terms of the Dottie number.
