Although the harmonic series $1 + \frac12 + \frac13 + \frac14 + \cdots$ diverges, we know that if we remove from the sum all the terms whose denominator expressed in base 10 contains a 9 digit, the series will converge.

Furthermore, the series will converge for any omitted digit (Kempner series).

I stumbled upon a book exercise that asks me to prove the convergence of every sum with removed digits for any numerical base b, for example:

Will the following sum:$1+\frac13 +\frac17 +\frac1{15}+\cdots$ (where all the positive integers that do not contain ”0” in the base 2 were removed) converge?

How can I prove it? Any help will be appreciated.(Sorry for my English)

  • 1
    $\begingroup$ The sum is majored by $1+\frac12 + \frac14 + \ldots = 2$. $\endgroup$
    – AlexR
    May 5, 2015 at 20:30
  • $\begingroup$ Do you know how to prove it in base $10$? The idea is the same for an arbitrary base. $\endgroup$ May 5, 2015 at 20:30

1 Answer 1


Cauchy's condensation test is enough for proving convergence.

For instance, between $2^N$ and $2^{N+1}-1$ there is just one number (the last one) without zeroes in its binary representation, and

$$ \sum_{n\geq 1}\frac{1}{2^n-1}$$ is quite trivially convergent. Another example: let $S=\{1,3,\ldots\}=\{s_1,s_2,\ldots\}$ be the set of positive integers for which the representation in base $7$ is free of $2$s. Between $7^N$ and $7^{N+1}-1$ there are $5\cdot 6^N$ elements of $S$, hence: $$ \sum_{n\geq 1}\frac{1}{s_n}\leq\sum_{m\geq 0}\frac{\left|[7^m,7^{m+1}-1]\cap S\right|}{7^m}=\sum_{m\geq 0}\frac{6^m}{7^m}=7.$$

  • $\begingroup$ but how can I prove the general case for any base? $\endgroup$
    – mobzopi
    May 5, 2015 at 22:04
  • $\begingroup$ @mobzopi: in the same way, you have just to adjust the numbers ($7$ for the base and $2$ for the non-allowed digit). $\endgroup$ May 5, 2015 at 23:03
  • $\begingroup$ sorry but what does # mean? $\endgroup$
    – mobzopi
    May 5, 2015 at 23:17
  • $\begingroup$ @mobzopi: cardinality of. $\endgroup$ May 6, 2015 at 11:03

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