I was just messing around in desmos when I came across this interesting function: $$f(x)=\sum_{n=0}^\infty \prod_{k=1}^n \frac{1}{k^x}$$
You can see it has a couple of interesting features, including asymptotes at x=0 and y=2, and its value at 1 is e.
I’m wondering if this has a closed form. It appears like a smooth function but I’ve never worked with infinite products.
Here is a closed form. First use product notation properties:
$$\sum_{n=0}^\infty \prod_{k=1}^n \frac{1}{k^x}=\sum_{n=0}^\infty \left(\prod_{k=1}^n k\right)^{-x}$$
then the definition of the factorial and the Pochhammer symbol $(a)_b$: $$\sum_{n=0}^\infty \left(\prod_{k=1}^n k\right)^{-x} =\sum_{n=0}^\infty \frac1{n!^x}=\sum_{n=0}^\infty \frac{1}{(1)_n^{x-1}n!}$$
Now use the Hypergeometric function $\,_p\text F_q$:
$$f(x)=\,_0\text F_{x-1}(\underbrace{1,…,1}_{x-1\text{ times}};1),x\in\Bbb N $$
See values of the function here
There are unfortunately only $4$ “special cases” with one using the Bessel I function from this result:
$$f(0)\to\infty$$ $$f(1)=e$$ $$f(2)=\text I_0(2)$$ $$f(\infty)\to2$$
How about for $x\not\in\Bbb N$?