Room isn't enough for 2

Let us divide the sum $S$ into a different sub-series we'll call $b_n$ , such that $b_n$ is the sum of all elements of $S$ of denominator $k$ for $10^{n-1} \leq k < 10^n$ .

So from this, we can get that

$b_1 = 1 + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{9}$ $b_2 = \frac{1}{10} + \frac{1}{11} + ... + \frac{1}{99}$

and so on.

Now we investigate the number of terms in each of $b_n$ .

It will be easy to note that $b_1$ has less than $9$ terms, or that $b_1 <9$ .

Similarly, we can also see that $b_2$ has less than $9^2$ terms, and that $b_2 < \frac{9^2}{10}$ .

In general, $b_n$ will have less than $9^n$ terms, and $b_n < \frac{9^n}{10^{n-1}}$ .

From here we can deduce that

$S < 9 + \frac{9^2}{10} + \frac{9^3}{10^2} + ...$

with which the RHS depicts a geometric series with common ratio $\frac{9}{10}$ .

So, we can say that $S < 90$ . Thus, the series is convergent .

*Notes:

• $S \neq 90$ .

• Fun fact : This also works for any series of the same format, but with any other digit removed (say, $6$ , or $7$ ).

Fix a positive integer $d$ , and consider all $k$ satisfying the given condition having exactly $d$ digits. There are $8 \cdot 9^{d-1}$ such $k$ (the first digit of $k$ has 8 choices: all except 0 and 2; the other digits have 9 choices: anything except 2). All these $k$ , being having $d$ digits, must be at least $10^{d-1}$ . Thus all these $k$ contribute at most $\dfrac{8 \cdot 9^{d-1}}{10^{d-1}}$ to the sum; each of these $k$ 's contribute $\frac{1}{k} \le \frac{1}{10^{d-1}}$ , and there are $8 \cdot 9^{d-1}$ of them. Summing over all $d$ , we have that the sum $S$ satisfies

$\displaystyle\begin{aligned} S &\le \sum_{d=1}^\infty \frac{8 \cdot 9^{d-1}}{10^{d-1}} \\ &= 8 \cdot \sum_{d'=0}^\infty \frac{9^{d'}}{10^{d'}} \\ &= 8 \cdot 10 \\ &= 80 \end{aligned}$

Thus $S$ has an upper bound. Since all terms of $S$ are increasing, by monotone convergence theorem $S$ converges .

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This can be generalized to ruling out any finite sequence of decimal digits. For example, if we take out all terms having the sequence $2017$ , the series will also converge.

This is called a Kempner Series .