A problem on a sequence I have this question: 


  Let $1,2,3,4,5,6,7,8,9,11,12,...$ be the sequence of all the positive integers which do not contain the digit zero. Write $\{a_n\}$ for this sequence. By comparing with a geometric series, show that $\sum_n \frac {1}{a_n} \lt 90$.


For the first part, I have found out the following steps to find $a_n$ when $n$ is given : 

$1.$ First we see if there exists some positive integer $k$ such that $n= \frac {9^{k}-9}{8}$. If such a $k$ exists, then 
$$a_n = 99 \cdots 99$$ where there are $k-1$ nines. 

$2.$ If not, then we compute the integer $m= \lfloor log_9 (8n+9)\rfloor$ and write the integer $n- \frac {9^{m} - 9}{8}$ in nonary system. Suppose the representation looks like $(b_m b_{m-1} \cdots b_2 b_1 )_9$. If $b_r$ is the last non-zero digit, then 
$$a_n = (b_m +1)(b_{m-1} +1) \cdots (b_{r+1} +1)(b_r)99 \cdots 99$$ where $(b_m +1)$ and $(b_{m-1} +1)$ and the similar terms are understood to be digits and there are $r-1$ nines. 
Now my questions are: 
$1.$ What does "Write $\{a_n\}$ for this sequence" mean? 
$2.$ Is there any better method to get $a_n$? Is there a closed form for $\{a_n\}$? 
$3.$ How to solve the second part of the question?
There are precisely $9^d$ $d$-digit numbers which do not contain the digit $0$. Each such $d$-digit number is clearly greater than $10^{d-1}$. Hence we have: $$\sum_{n\in\mathbb{N}}\frac{1}{a_n}&lt;\sum_{d\in\mathbb{N}}\frac{9^d}{10^{d-1}} = 90$$
In fact, we always have that

$$\sum\limits_{n \in \mathbb{N}} \frac{1}{a_n} =(\sum\limits_{n \in \mathbb{N}} \frac{1}{n}) - (\sum\limits_{n \in \mathbb{N}} {\frac{1}{10 . n}}) = \frac{9}{10} (\sum\limits_{n \in \mathbb{N}} \frac{1}{n}).$$