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Update: This question has been resolved. I have made some mistakes in this post. I will leave my post here for readers to find out my mistakes.


I have noticed that the post is a bit too long. So if you don't bother to read, here is the question:

By density I mean the proportion of the number of terms we counted in the sum to the total number of terms, as the number of terms approaches infinity. You can find examples down in my post.

My theory is that, in order for the sum of a subset of the harmonic sequence to converge, its density must be $0$. Is it possible to prove or disprove it? Or is my definition for density consistent? In case it is false, could anyone come up with a counter-example?


Background:

So I recently read about Kempner's series and it is counter-intuitive at first glance. Based on it and a number of other examples, I have come up with a theory which I am not able to prove or disprove. But first, let's look into the background:

Kempner's Series

Suppose you have the harmonic series, which is divergent. But instead of summing up all the terms, we omit every terms containing any specific finite string of numbers. The series will always converge. For example, removing all numbers that contain a $9$ in its digit:

$$K_9=1+\frac 12 +...+ \frac 18 + \frac 1{10}+...+\frac 1{18}+\frac 1{20}+...\approx 22.92$$

The fascinating fact of the Kempner's series is that no matter how long the string of numbers you choose, the series will always converge. So I look into it to try to understand what is going on.

As it turns out, the number of terms of that contain a specific string will be fewer as the number gets big. Using the above example, between $1$ to $10$, we remove only $1$ term. Up to $100$, we remove $1*9+10=19$ terms. Up to $100$, we remove $19*9+100=271$ terms, etc. Let $A_n$ be the number of terms that are removed below the first $10^n$ terms in the harmonic sequence. We get

$$A_n=9A_{n-1} +10^{n-1}$$

Where $A_1=1$

So,

$$A_n=10^n-9^n$$

Define the density of the sequence as

$$\rho=\lim_{n\to \infty} \frac{N(n)}{n}$$

Where $N(n)$ denotes the number of terms counted from the harmonic sequence up to the $n^{th}$ harmonic number.

In our case,

$$\rho =1- \lim_{n\to \infty} \frac{10^n -9^n}{10^n} = 0$$

Which means the proportion of terms counted actually approaches to $0%$, which is a non-rigorous argument I found to show that the sum converges. Since I was not sure if this idea works, so I looked into more cases.


The sum of reciprocals of all prime numbers:

Using the approximation $\pi (x) \approx \frac{x}{\log x}$, it can be shown easily that the density of terms removed approaches 0, as:

$$\rho =1-\lim_{x\to \infty} \frac 1{\log x} =1$$

Which matches with my conjecture above because the sum diverges.


Another case: sum of any harmonic progression is divergent, i.e. for any positive integer $a$ and $d$

$$\sum_{k=0}^{\infty} \frac 1{a+kd} \to \infty$$

It is easy to see the density of terms counted is $\frac 1d$ as we are only counting a term for every $d$ terms. So it matches with my conjecture above.


Another example could be any geometric progression of the type:

$$\sum_{k=0}^{\infty} x^{-k}=\frac x{x-1}$$

For any positive integer $x>1$, the sum converges. Because we only count a term for every $x^k$, which means we count less and less terms as $k$ increases, the density matches my theory again:

$$\rho = 1-\lim_{n\to \infty} \frac {n+1}{x^n}=0$$


Questions: As my mathematical background is not strong, I have no idea about whether this can be proven or not. Or if this conjecture happens to be false, can anyone show me a counter-example? Thanks.

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It's a nice idea, but unfortunately you found a counterexample yourself, you just misinterpreted it.

You write: "As it turns out, the number of terms of that contain a specific string will be fewer as the number gets big." It's actually the other way around, and you show this in your calculations: The proportion of terms that contain a specific string goes to $1$, and the proportion of terms that are left goes to $0$.

The proportion of primes also goes to $0$. So this is in fact the opposite case, and a counterexample to your conjecture. Despite the density going to zero, the sum of the reciprocals of the primes diverges.

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  • $\begingroup$ Oh! You are absolutely right! Haha, I was so blind that it is embarrassing :) Thanks man! $\endgroup$
    – lEm
    Apr 7 '16 at 15:27

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