How can i prove that this proposition?

Question

Let S be a set of natural number that does not contain the digit 7 in their decimal representation.

prove that

$\sum_{n\in S}\frac{1}{n}$ < ∞

help me please

Is that mess meant to $\sum_{n\in S}\frac{1}{n}$ ? – almagest Apr 10 '16 at 14:28 — almagest, Apr 10 '16 at 14:28
@almagest yes.. i have some trouble in texting words. sorry – user330235 Apr 10 '16 at 14:41 — user330235, Apr 10 '16 at 14:41

score 1 · Accepted Answer · answered Apr 10 '16 at 15:22

The idea is that most of the natural numbers actually have a 7 in their decimal expansion.

More specifically, let $p$ be an integer, we know that $\forall n,10^p\leqslant n < 10^{p+1}\Rightarrow 10^{-(p+1)}<\frac 1 n \leqslant 10^{-p}$.

Now between $10^p$ and $10^{p+1}$ there are exactly $8\times 9^{p-1}$ numbers that don't have 7 in their decimal expansion (simple combinatorics).

So $\sum_{n\in\mathcal S}\frac 1 n= \sum_{p\in \mathbb N}\sum_{n=10^p,n\in\mathcal S}^{10^{p+1}-1}\frac 1 n\leqslant \sum_{p\in \mathbb N} 8\frac {9^{p-1}}{10^p}<\infty$

By the way this works for every number from 0 to 9, not only for 7.

How can i prove that this proposition?

1 Answers1