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Does the following series converge or diverge? $\sum_{n=1}^{\infty} a_n$ where $a_n = \frac 1 b_n$, and $(b_n)_n$ is the subsequence of $(n)_n$ whose terms do not have a $9$ in their decimal notation.

I have no idea how the sequence $b_n$ even looks like. If I have an idea of how to prove that a series converges or not with criteria but I don't if I don't know what $b_n$ is.

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  • $\begingroup$ $b_n$ starts $1,2,3,4,5,6,7,9,10,11,12,13,14,15,16,17,19,\dotsc,86,87,99,100,\dotsc$, if I interpret "followers" right. $\endgroup$
    – Daniel Fischer
    Commented Nov 27, 2013 at 14:28
  • $\begingroup$ What do you mean by "followers"? It seems $(b_n)$ is just a sequence of natural numbers of which no $b_{n+1}$ contains $9$ in it's base $10$ expansion. for example, 179 is in $(b_n)$ but 178 is not in $(b_n)$. $\endgroup$
    – doppz
    Commented Nov 27, 2013 at 14:28
  • $\begingroup$ i think its more like 1,2,3,4,5,6,7,8,10...,17,18,20,..87,88,100 cause numbers like 19 or 99 have a 9 in there decimal notation $\endgroup$
    – Weenzeal
    Commented Nov 27, 2013 at 14:31
  • $\begingroup$ @user111849 The question is how to interpret "whose followers don't have a 9 in their decimal representation". I interpret it as $n$, such that the decimal representation of $n+1$ contains no $9$. $\endgroup$
    – Daniel Fischer
    Commented Nov 27, 2013 at 14:34
  • $\begingroup$ i think i used the wrong word with followers but I didn't found another one :/ word to word translation would be sequence items. $\endgroup$
    – Weenzeal
    Commented Nov 27, 2013 at 14:37

2 Answers 2

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Let $c_n$ be the sum of all reciprocals of $n$-digit positive numbers that contain no '9' digit. There are $8 \cdot 9^{n-1}$ $n$-digit numbers that contain no '9' digit. The largest term is the first one, $\dfrac{1}{10^{n-1}}$. So $c_n < 8 \cdot 9^{n-1} \dfrac{1}{10^{n-1}} = 8 \left(\dfrac{9}{10}\right)^{n-1}$.

You're looking for $\sum c_n$, which is smaller than the geometric series $\sum 8 \left(\dfrac{9}{10}\right)^{n-1}$, so $c_n$ certainly converges.

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    $\begingroup$ very elegant bound $\endgroup$
    – Charlie Tian
    Commented Dec 14, 2017 at 15:36
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It does converge, it is a matter of counting the right thing:

First you notice that powers of ten will tell you how much numbers have to be seperate from the natural numbers.

10-> 1 number (9) 100-> 19 numbers (9,19,...,89,90,91...99) 1000-> 19*9 numbers (19 from 101 to 199, etc.. 801 to 899) + 100 (900,901...999) etc..

So if you let $D_k$ be the number of integers smaller than $10^k$ and with the number 9 in their decimal notation, you have: D1 = 1 D2=9*1 + 10 D3= 9*19 + 100

You induce the formula: $D_{k+1} = 10^k +9*D_k$ (which you can prove using a recurrence)

To solve this you divide by $9^{k+1}$, a telescoping appears and you can easily calculate the value of $D_k$. $D_k = 10^k - 9^k$

Of course you could have found this with a direct counting.

Now : It means that there is $10^k - (10^k -9^k) = 9^k$ numbers without a nine in their decimal notation between zero and $10^k$, aka : $B_{9^k} = 10^k$ (the $9^k$-th term of Bn is $10^k$)

Then, you are going to sum the $(B_n)$ by blocks of $9^k$ +1 to $9^{k+1}$ :

n part of [$9^k$ +1, $9^{k+1}$] , $B_n >= B_{9^k} = 10^k$ --> $A_n =< 10^{-k}$

--> $S_k = \sum_{n =9^k +1}^{9^{k+1}} A_n \leq \frac{9^{k+1}-9^k}{10^k} = 8*(\frac{9}{10})^k$

So the series of general term $(S_k)$ converges by comparison with a geometrical series, hence you have the convergence of the series of general term $(A_n)$.

Thx to those who read it all along.

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