Better upper bound for $ \sum_{k=2}^{\infty} \frac{1}{2^{k-1}} \sum_{n=1}^{\infty} (n! \: \text{mod} \: k) $ Since the sum $\sum_{n=1}^{\infty} (n! \: \text{mod} \: k)$ will be zero beyond $k-1$, the series could be interpreted as an finite sum of length $k-1$. Also, the max value of $n! \: \text{mod} \: k$ is naturally $k-1$.
Taken together one has a max value of:
$$ \sum_{k=2}^{\infty} \frac{(k-1)^2}{2^{k-1}} = 6 $$
Mathematica gives the value of
$$ \sum_{k=2}^{\infty} \frac{1}{2^{k-1}} \sum_{n=1}^{\infty} (n! \: \text{mod} \: k) \approx 3.005674093 $$ so my upper bound is clearly quite crude. What would be a better upper bound?
Edit: To be clear my upper bound is $\frac{x (x+1)}{(1-x)^3}$ for $ \sum_{k=2}^{\infty} x^{k-1} \sum_{n=1}^{\infty} (n! \: \text{mod} \: k)$.

Calculate more terms we could reach more accurate bound.
Since $\sum_{k=n}^{\infty}\frac{(k-1)^2}{2^{k-1}}=
\frac{(n-1)^2}{2^{n-4}}-\frac{2n^2-6n+3}{2^{n-3}}+\frac{(n-2)^2}{2^{n-2}}$
So we have the sum is between $\sum_{k=2}^{h-1}\frac1{2^{k-1}}\sum_{n=1}^{\infty}n!\pmod k$ and $\left(\sum_{k=2}^{h-1}\frac1{2^{k-1}}\sum_{n=1}^{\infty}n!\pmod k\right) +\frac{(h-1)^2}{2^{h-4}}-\frac{2h^2-6h+3}{2^{h-3}}+\frac{(h-2)^2}{2^{h-2}}$.
For example when h=10, we have the result is between $\frac{93}{32} \approx2.91$ and $\frac{423}{128}\approx3.30$; when h=20, the result is between $\frac{393939}{131072}\approx3.0055$ and $\frac{98535}{32768}\approx3.0070$.
