Problem 368

A Kempner-like series

The harmonic series $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ … $is well known to be divergent.

If we however omit from this series every term where the denominator has a 9 in it, the series remarkably enough converges to approximately 22.9206766193.
This modified harmonic series is called the Kempner series.

Let us now consider another modified harmonic series by omitting from the harmonic series every term where the denominator has 3 or more equal consecutive digits. One can verify that out of the first 1200 terms of the harmonic series, only 20 terms will be omitted.
These 20 omitted terms are:

$\frac{1}{111}$, $\frac{1}{222}$, $\frac{1}{333}$, $\frac{1}{444}$, $\frac{1}{555}$, $\frac{1}{666}$, $\frac{1}{777}$, $\frac{1}{888}$, $\frac{1}{999}$, $\frac{1}{1000}$, $\frac{1}{1110}$,
$\frac{1}{1111}$, $\frac{1}{1112}$, $\frac{1}{1113}$, $\frac{1}{1114}$, $\frac{1}{1115}$, $\frac{1}{1116}$, $\frac{1}{1117}$, $\frac{1}{1118}$ and $\frac{1}{1119}$.

This series converges as well.

Find the value the series converges to.
Give your answer rounded to 10 digits behind the decimal point.

类肯普纳级数

众所周知，调和级数 $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ … $是发散的。

不过，如果我们忽略所有分母中含有数字9的项，新的级数将神奇地收敛到大约22.9206766193。
这种修正调和级数被称为肯普纳级数。

现在让我们考虑另外一种修正调和级数，忽略所有分母中有至少3个连续相同数字的项。可以验证，在调和级数的前1200项中，只有20项会被忽略。
这20个被忽略的项是：

$\frac{1}{111}$、$\frac{1}{222}$、$\frac{1}{333}$、$\frac{1}{444}$、$\frac{1}{555}$、$\frac{1}{666}$、$\frac{1}{777}$、$\frac{1}{888}$、$\frac{1}{999}$、$\frac{1}{1000}$、$\frac{1}{1110}$、
$\frac{1}{1111}$、$\frac{1}{1112}$、$\frac{1}{1113}$、$\frac{1}{1114}$、$\frac{1}{1115}$、$\frac{1}{1116}$、$\frac{1}{1117}$、$\frac{1}{1118}$和$\frac{1}{1119}$。

新的级数同样收敛。

求这个级数的收敛值。
将你的答案四舍五入到小数点后10位数字。