# Proof that the harmonic series is < $\infty$ for a special set..

In one of my books i found a very interesting task, i am really curios about the solution:

Let $M = \{2,3,4,5,6,7,8,9,20,22,...\} \subseteq \mathbb{N}$ be a set that contains all natural numbers, that don't contain a "1" in their depiction. Show that:

$$\sum_{n\in M} {\frac{1}{n}} < \infty$$

The Set M that we created is infinite right? I will always be able to find a natural number not containing a "1" that is larger. So isn't it kind of the same deal, like the harmonic series, from which we know it diverges?

Sure,$\frac{1}{n}$ will converge "quicker" against 0 so the series might converge then.. but i find that kind of hard to prove.

How "quick" does the sequence 1/n have to converge against 0, so that our series is convergent? What do we need to be able to say about the partial sums? Is there a convergence criteria applicable here?

(I think $<\infty$ means it has to converge at some stage right?)

I also read the proof that the "normal" harmonic series is divergent. It estimates the partial sums (I can provide this proof if necessary)... But in this case, what can you say about $M$ to lead a proof like this to a contradiction?

I got such an foggy conception of this, if somebody could shed some light on this that would be awesome!

• This is a Kempner series. See answers on related questions here, here and there. Commented Nov 25, 2014 at 17:31
• That is a good hint thank you. Commented Nov 25, 2014 at 17:33

Begin by defining

$$S(1;9) := \sum_{i=2}^9 \frac{1}{i}$$

Let $a = S(1;9)$. Then $a < 2$. Let $S(10;99)$ be the sum over the fractions of all numbers between $10$ and $99$ that don't contain a $1$ in their depiction.

Prove that $S(10;99) < \frac{9}{10}a$. Similarly, define $S(100;999)$ to be the sum over the fractions of all numbers between $100$ and $999$ that don't contain a $1$ in their depiction. Prove that $S(100;999) < (\frac{9}{10})^2a$ etc etc.

The sum we are looking for is

$$S(1;9) + S(10;99) + S(100;999)...$$

But now we can easily give an upper bound for this sum.

• Using the very same method, you can, in fact prove, that this sum converges for any fixed sequence of digits that is not allowed to appear in the depiction of these numbers. Commented Nov 24, 2014 at 22:34
• Now generalize this to prove the Erdos Turan conjecture. Just kidding :). Commented Nov 24, 2014 at 23:06
• But how can i prove that $S(10;99) < \frac{9}{10}a$, How does this function S look for S(10,99)? An idea i had was looking at the partial sums of the "normal" harmonic series, and then claiming that our partial sums have to be smaller. Commented Nov 25, 2014 at 17:22
• after going through the links from @achille hui I get you now. Thanks for your effort. Commented Nov 25, 2014 at 21:11