How to sum this series:
$$\frac{1}{1}+\frac{1}{11}+\frac{1}{111}+\frac{1}{1111}+\cdots$$
My attempt:
Multiply and divide the series by $9$
$$9\left(\frac{1}{9}+\frac{1}{99}+\frac{1}{999}+\frac{1}{9999}+\cdots\right)$$
$$9\left(\frac{1}{10-1}+\frac{1}{10^2-1}+\frac{1}{10^3-1}+\frac{1}{10^4-1}+\cdots\right)$$
Now let $a_N$ denote the number of divisors of $N$, after some simplification the series becomes:
$$9\left(1+\sum{\frac{a_N}{10^N}}\right)$$
This is where I am stuck...
PS: Please rectify my mistakes along the way
Answer
Your approach is very nice but, as pointed by Pranav Arora, for the summation up to term $n$, a CAS leads to $$S =9 \left(\frac{\psi _{\frac{1}{10}}^{(0)}(n+1)}{\log (10)}-\frac{\psi
_{\frac{1}{10}}^{(0)}(1)}{\log (10)}\right)$$ and for the infinite summation, it becomes $$S=\frac{9 \left(\log \left(\frac{10}{9}\right)-\psi
_{\frac{1}{10}}^{(0)}(1)\right)}{\log (10)} \simeq 1.100918190836200736379855$$
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