Formula to find number of divisors of a number Consider the sum $\sum_{n=1}^{k}(\frac{1}{10^n-1}) = 0.122324243426 = a$. Where k is an arbitrary range we choose.
We can see that the decimal place of $10^{-n}$ is the number of divisors of n. This occurs because the sum basically acts like a sieve of eratosthenes, but within the decimal. 
To find the value at $10^{-n}$ we can use :  $\lfloor\operatorname{mod}\left(a\cdot10^{x},10\right)\rfloor$
The issue with this is that it will work only if all the numbers less than $k$ have less than 10 divisors. To fix this we can just use a larger base and repeat the process, for example 100. 100 would work for $k < 10000$. 
This seems like an obvious property and is even a direct formula as long as you have the exact value for $a$. Does this count as a closed form for the divisor function $\sigma_0(n)$ for $n<k$?  
 A: What you are describing is the Lambert series generating function of the number of divisors of $n$ as given in the first example in the Wikipedia article. This is the
OEIS sequence A000005 which also gives its generating function as
$$ \sum_{n=1}^\infty \sigma_0(n) x^n = \sum_{k=1}^\infty x^k/(1 - x^k)
 = \sum_{k=1}^\infty 1/(y^k - 1) \text{ where } y := 1/x.$$
As you noticed, if $y$ is a power of ten, then in the decimal expansion of the infinite sum the sequence appears up to the point where the sequence terms are bigger than $y$. The question of if this is a closed form for $\sigma_0$ depends entirely on the precise definition of "closed form" but there are several possible definitions of the term.
By  the way, the same idea work for other integer sequences and their generating functions. A famous example is the Fibonacci numbers sequence A000045
$$ \sum_{n=1}^\infty F_n x^n = x/(1 - x - x^2)
 = y/(y^2 - y - 1) \text{ where } y := 1/x.$$
If $y=10$ then we have $10/(10^2-10-1) = 10/89 \approx 0.1123595505$ where the first few terms of $F_n$ appear in the decimal expansion.
