Does this weird series converge? $\sum_{n\in S}$$\frac{1}{n}$, where S consists of those positive integers whose decimal expansion does not contain the digit 1.
This was a part(b) question.  
Part (a) was an evaluation of the double integral $\int \int_S$ $(x^2 + y^2)d\sigma$, where S is the unit sphere centered at (0,0,0), and $\sigma$ is surface area.  The answer is $\frac{8\pi}{3}$.
I don't see how part(a) can be used to solve part(b).  
For the series evaluation, if we are summing over the positive integers that don't contain the digit 1, then I'm thinking we exclude integers such as 1, which has decimal expansion .99999.. or 1.0000, or 10, which has expansion 9.9999... or 10.0000, etc, and we sum over numbers such as 2, 3, 4, 5, ... 
Not sure where I can go with this idea, though.
Thanks,
 A: Let me reformulate my answer. Now let $S$ be the subset of $\mathbb N$ whose decimal expansion do not contain digit $1$.
The answer is the series $\sum_{n\in S}\frac{1}{n}$ is convergent. 
 In contrast, think about the fact $\sum_{p\in prime}\frac{1}{p}=\infty$, and the sparsity of primes when they are large. It shows that $S$ occupies a extremely "meager" set in  $\mathbb N$.
Before we come to the proof, here are two observations:
(1)Think of this question: If we take decimal expansion of each real $x\in [0,1]$, $x=\alpha_1\alpha_2...\alpha_n...$ , where $\alpha_i\in \{0,1,..,9\}$, then remove those reals whose expansion ever contains digit $1$, what set will we get？ The answer is we will get a measure zero set, since it is a variant of Cantor set whereas the Cantor set takes ternary expansion. By this observation, we have a strong feeling that $S$, those integer do not contain digit $1$, only occupies a "measure zero" set in $\mathbb N$.
(2)Let $T_n$ be the set of integer contains only $n$-digit integer, i.e. $$T_n=\{k\in \mathbb N| 10^n\le k\le10^{n+1}-1\}.$$ and let $S_n=T_n \cap S.$ So $S=\cup_{n\in \mathbb N}S_n.$ We have $${{|S_n|}\over{|T_n|}}={{C_8^1(C_{9}^1)^{n-1}}\over {C_9^1(C_{10}^1)^{n-1}}}={{8\times 9^{n-1}}\over{9\times 10^{n-1}}}\le ({{9}\over{10}})^n$$
The proportion of $|S_n|$ in $|T_n|$ will be small when $n$ goes large.
Finally, let's prove the claim that $\sum_{n\in S}\frac{1}{n}$ is convergent, notice that each $n\in S_n, n\ge 10^n.$ Then,
$$\sum_{n\in S}\frac{1}{n}=\sum_{n\in\mathbb N}\sum_{k\in S_k}\frac{1}{k}
\le\sum_{n\in\mathbb N}\sum_{k\in S_k}\ ({{1}\over{10}})^n
\le\sum_{n\in\mathbb N}({{9}\over {10}})^n\times ({{1}\over{10}})^n
=\sum_{n\in\mathbb N}({{9}\over{100}})^n
=\frac{100}{91}.$$
So the series is convergent.
A: For part (b), $S$ has $8$ one digit integers, $8 \cdot 9$ two digit integers, $8 \cdot 9^2$ three digit integers, etc. 
So, the sum $\displaystyle\sum_{n \in S}\dfrac{1}{n}$ has $8$ terms that are at most $\dfrac{1}{2}$, $8 \cdot 9$ terms that are at most $\dfrac{1}{20}$, $8 \cdot 9^2$ that are at most $\dfrac{1}{200}$, etc. 
Can you do something with this? The formula for the sum of a geometric series will also be helpful.
I'm also not sure how parts (a) and (b) are related. It's entirely possible that they are not.
