# A modification of the harmonic series that causes it to converge. [duplicate]

Possible Duplicate:
sum of inverse of numbers with certain terms omitted

We are working in base $b$, a positive integer greater than $1$. Let $S$ be the set of all positive integers that contain a given symbol in their base $b$ expansion. Let's choose the symbol to be $1$, (because every base we are considering uses this symbol). Show that the following series converges: $$\sum_{n\in({\mathbb{Z}^+}-S)}^\infty \frac{1}{n}$$ For example, if we look at base 10, the sum would start like $$\frac{1}{2}+\frac{1}{3}+...+\frac{1}{9}+\frac{1}{20}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{30}+\frac{1}{32}+...$$

Full disclosure: I got this question from my math class. I have a solution to it (largely inspired by the methods of others in my class), which I will post after awhile. I'm interested in seeing different solutions, as is discussed here http://meta.math.stackexchange.com/questions/4232/is-it-okay-to-ask-mathematical-puzzles-and-problems-i-have-solved

-

## marked as duplicate by Ｊ. Ｍ., Matt N., Old John, Jonas Teuwen, Pedro TamaroffAug 19 '12 at 3:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

You should really be accepting the answer that you liked or found more useful, I think, not the one that others think is useful. Votes here are sometimes misleading. –  Pedro Tamaroff Aug 18 '12 at 16:35
Ok, that's fine. I'll remove that statement from the post. –  C. Williamson Aug 18 '12 at 16:37
Your series is called a Kempner series. See this question. –  Ｊ. Ｍ. Aug 18 '12 at 17:36
See also this article and this article. –  Ｊ. Ｍ. Aug 18 '12 at 17:40
@J.M. great find! This is cool because I was working on proving that if one excludes any sequence of length n from appearing in the number, then the series still converges. –  C. Williamson Aug 18 '12 at 20:46

## 2 Answers

There are $(b-2)(b-1)^{k-1}$ elements $n$ which aren't in $S$ and have exactly $k$ digits, and for each of these you have $\frac{1}{n} \le \frac{1}{b^k}$. So you have an absolutely convergent majorant: $$\sum_{n \notin S} \frac{1}{n} \le (b-2)\sum_{k=1}^{\infty} (\frac{b-1}{b})^k < \infty$$

-
Very nice and succint. You'll see that our methods are essentially the same. I wonder if there is another "good" way to solve this. –  C. Williamson Aug 18 '12 at 16:49
@C.Williamson I'm interested as well. –  Cocopuffs Aug 18 '12 at 16:50

First, consider numbers $n$ that are in the range $[b^k, b^{k+1})$. Since there are $b-1$ choices for the first digit in the base $b$ representation of $n$ (as zero cannot be leading), $\frac{b-2}{b-1}$ is the ratio of numbers that do not contain a $1$ in the first position. So, these numbers are the only ones still in $(\mathbb{Z}^+-S)$ and therefore the only ones that may still end up in the terms of the series. However, there is still a $\frac{1}{b}$ chance for each of the remaining positions in that number that a $1$ will appear.

Therefore, in the range $[b^k, b^{k+1})$, there are $(b^k-b^{k+1})(\frac{b-2}{b-1})(\frac{b-1}{b})^k$ numbers that will not be in set $S$ and thus have their reciprocal added to the series. Note also that every term in this range must be less than or equal to $\frac{1}{b^k}$. Therefore, the sum of the terms in the range must be less than or equal to the number of terms present in the range times $\frac{1}{b^k}$: $$\sum_{terms\in{[b^k, b^{k+1})}}(terms)^{-1}\leq{(b^k-b^{k+1})\left(\frac{b-2}{b-1}\right)\left(\frac{b-1}{b}\right)^k\left(\frac{1}{b^k}\right)}$$ It follows that: $$\sum_{n\in({\mathbb{Z}^+}-S)}^\infty \frac{1}{n}=\sum_{k=0}^\infty \sum_{terms\in{[b^k, b^{k+1})}}(terms)^{-1} \leq{\sum_{k=0}^\infty\left[ (b^k-b^{k+1})\left(\frac{b-2}{b-1}\right)\left(\frac{b-1}{b}\right)^k\left(\frac{1}{b^k}\right)\right]}$$ The last summation can be simplified to $$\left(\frac{b-2}{b-1}\right)\sum_{k=0}^\infty\left[(b-1)\left(\frac{b-1}{b}\right)^k\right] = (b-2)\sum_{k=0}^\infty\left(\frac{b-1}{b}\right)^k$$. This is a geometric series, which converges to $\frac{b-2}{1-\frac{b-1}{b}} = b(b-2)$. So, we have: $$\sum_{n\in({\mathbb{Z}^+}-S)}^\infty \frac{1}{n}\leq{b(b-2)}<{\infty}$$ We have just seen that the removal of all terms that contains one or more symbol in base $b$ from the harmonic series causes the series to converge in every base. Intuitively, the larger the base, the more options we have. This would mean we have fewer terms removed and a larger sum, which is suggested but not proved by the final inequality. In base 2, there are so few options (none, in fact) that the series has a value of 0 (which is less than or equal to $b(b-2)$).

-