Strange Factorial Sum

[This is not a full solution because I'm too lazy to write one ^_^]

We see that the taylor polynomial of $\dfrac{\sin x+\cos x+e^x}{2}=1+x+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+\dfrac{x^8}{8!}+\cdots$

Plugging in $x=1$ gives $\dfrac{\sin 1+\cos 1+e^1}{2}=2+\dfrac{1}{4!}+\dfrac{1}{5!}+\dfrac{1}{8!}+\cdots$

We subtract $2$ to get the sum we are looking for.

Thus the exact value is $S= \dfrac{\sin 1+\cos 1+e}{2}-2$

where we are working in radians.

We have that $S=\dfrac{\sin 1+\cos 1+e}{2}-2\approx 0.050$ so $\lfloor 1000S\rfloor = \boxed{50}$

This can be very easily found using partial sums, as even only $\dfrac{1}{4!}+\dfrac{1}{5!}$ gives $S=0.05$ . However I hope that people will try to find the exact value $\ddot\smile$

A classical way to extract some terms of the exponantial series is to use an $n^{th}$ root of unity which is not 1:

Let $\omega = e^{i \frac{2\pi}{n} }$

Then,

$e^x + e^{\omega x} + \cdots + e^{\omega^{n-1} x} = \displaystyle\sum_{k=0}^{+\infty} x^k\dfrac{1+\omega^k+\omega^{2k}+\cdots+\omega^{(n-1)k}}{k!}$

But $1+\omega^k+\omega^{2k}+\cdots+\omega^{(n-1)k} = \begin{cases} n \quad \text{if} \quad k\mod n=0 \\ 0 \quad \text{else} \end{cases}$

So $\quad \displaystyle\sum_{k=0}^{+\infty} \dfrac{x^{nk}}{(nk)!} = \dfrac{e^x + e^{\omega x} + \cdots + e^{\omega^{n-1} x}}{n}$

You can integrate to get $\displaystyle\sum_{k=0}^{+\infty} \dfrac{x^{nk+1}}{(nk+1)!}$

Applying this to $n=4$ , we get $\displaystyle\sum_{k=1}^{+\infty} \dfrac{x^{4k}}{(4k)!} + \dfrac{x^{4k+1}}{(4k+1)!} = \dfrac{\cos(x)+\sin(x)+e^x}{2}-1-x$

So the answer is $\boxed{\dfrac{\cos(1)+\sin(1)+e-4}{2}}$

Looking at the series, we group in a pair of two items., first two is n=1, second two n=2, third two n=3 gives $\frac{1}{12!} + \frac{1}{13!}$
1000 times the floor of these would be 0 + 0
Similarly all the rest will be zeros.
Thus the sum of the first given four items and its floor value is found to be $\boxed {50}$