TITLE: Interesting convergence in divisor sums up to $10^k$
QUESTION [11 upvotes]: 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 $$S'(\infty) = \frac{\pi^2}{12}$$?

REPLY [6 votes]: 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).