Let $S(k)$ be the sum of divisors across each of $1, 2, ..., 10^k$.
For example, $$\begin{align}S(1) &= 1 + 3 + 4 + 7 + 6 + 12 + 8 + 15 + 3 + 17 \\&= 87\end{align}$$
where each of $1 + 3 + ... + 17$ are the sums of divisors of $1, 2, ..., 10^1$ respectively.
I noticed that as $k$ increases, the decimal expansion of $S(k)$ seems to "converge" to some number. For the first few values of $k$, computed using some suitable script, we have
$$\begin{align} S(1) &=87\\ S(2) &=8299\\ S(3) &= 823081\\ ...\\ S(10) &= 82246703352400266400\\ ...\\ S(15) &=822467033424114009326065894639\\ S(16) &=82246703342411333689227187822414\\ S(17) &= 8224670334241132270081671519064067\\ \end{align}$$
If we take the decimal expansion of $S(k)$ as a fractional part and let this be $S'(k)$, we have
$$S'(\infty) = 0.82246703342411...$$
But hey!
$$\frac{\pi^2}{12} = 0.82246703342411...$$
So my question is this : Is the limit indeed $$S'(\infty) = \frac{\pi^2}{12}$$?
This is all due to the fact that $$ \frac{1}{x}\sum_{n\le x} \sigma(n) = \frac{\pi^2}{12}x + O(\log x). $$ See these notes by Carl Pomerance (equation (5) if you're in a hurry).