# Number theory question

## Homework Statement

Does the sum of the reciprocals of natural numbers starting with nine converge?
In other words, does Sigma 1/n with n being numbers starting with nine, converge?

## The Attempt at a Solution

I know that subsets of the natural numbers with positive density, their sums of reciprocals will diverge. I know that numbers beginning with nine have zero density. So my gut instinct is that is does converge.

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It will converge if an only if the summation
$$\sum_{x=N}^{\infty}\frac{1}{x}$$
satisfies the integral test.
The integral test states that the series only converges if an only if the integral
$$\int_{N}^{\infty}\frac{1}{x} dx$$
is finite. In this case you can evaluate it by using $$ln(x)$$

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HallsofIvy
Homework Helper
That would be relevant if he were summing 1/n for all n. But he specifically said he summing 1/n ONLY for n that has first digit 9. That's a completely different question.

I thought the integral test was satisfied when beginning with any $$N$$ in the harmonic series.

Dick
Homework Helper
I thought the integral test was satisfied when beginning with any $$N$$ in the harmonic series.
The series isn't all of the numbers after 1/9. It's all of the numbers that start with the digit '9'. I.e. 1/9+1/90+1/91+...+1/99+1/900+... lei123 should try to estimate the sum. I don't think the numbers starting with 9 have zero density. There's quite a lot of them.

Aha! I see.

I recall my professor specifically said that it had zero density. I agree that there are a lot of numbers starting with nine, but when you take the limit of ratio of the (# of #s starting with nine)/(infinity), it would seem to tend to zero.

also, lei123 is a girl ;)

Dick
Homework Helper
I recall my professor specifically said that it had zero density. I agree that there are a lot of numbers starting with nine, but when you take the limit of ratio of the (# of #s starting with nine)/(infinity), it would seem to tend to zero.

also, lei123 is a girl ;)
I think you have to define the density as (# of numbers starting with 9 that are less than N)/(# of numbers less than N) and take the limit as N->infinity. You don't divide by infinity. I'd take a guess and say that's about 1/9. But I could be a hair off. It's definitely not zero.

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All the numbers starting with nine are of the form:

$$k = 9*10^{m} + l$$

where $m$ and $l$ are non-negative integers. I think the convergence of the sum is the same as the convergence properies of the double integral:

$$\int_{0}^{\infty}{\int_{0}^{\infty}{\frac{dy dx}{9*10^{x} + y}}}$$

The integral w.r.t. y diverges logarithmically.

Correction, the $l$ has to be smaller than $10^{m + 1}$, otherwise it will contribute digits in front of the 9. This means that the limits of the above integral w.r.t. $y$ are from $0$ to $10^{x + 1}$. Then, you could have convergent integral, but you need to investigate it yourself.

Dick
Homework Helper
Correction, the $l$ has to be smaller than $10^{m + 1}$, otherwise it will contribute digits in front of the 9. This means that the limits of the above integral w.r.t. $y$ are from $0$ to $10^{x + 1}$. Then, you could have convergent integral, but you need to investigate it yourself.
There is a LOT simpler way to do this without any integral test which I don't even think you can comfortably apply. 1/9>1/10, 1/90+1/91+...+1/99>1/10. That's a hint for the OP. A very strong hint.

There is a LOT simpler way to do this without any integral test which I don't even think you can comfortably apply. 1/9>1/10, 1/90+1/91+...+1/99>1/10. That's a hint for the OP. A very strong hint.
Actually, your inequalities are messed up.

It follows from a different inequality:

1/90 + 1/91 + ... + 1/99 > 10/99

1/900 + 1/901 + ... + 1/999 > 100/999

...

and from investigating the series:

$$\frac{1}{9} + \frac{10}{99} + \frac{100}{999} + \ldots = \sum_{n = 1}^{\infinity}{\frac{10^{n - 1}}{10^{n} - 1}}$$

I think you need to use the Integral test for this one to prove its convergence properties.

Dick
Homework Helper
It follows from a different inequality:

1/90 + 1/91 + ... + 1/99 > 10/99

1/900 + 1/901 + ... + 1/999 > 100/999

...

and from investigating the series:

$$\frac{1}{9} + \frac{10}{99} + \frac{100}{999} + \ldots = \sum_{n = 1}^{\infinity}{\frac{10^{n - 1}}{10^{n} - 1}}$$

I think you need to use the Integral test for this one to prove its convergence properties.
10/99>1/10
100/999>1/10
etc. And if I were looking at a series whose terms don't approach zero, I wouldn't spend a lot of time on other convergence tests.

this is interesting - in my probability theory class today we learned about Benford's law - I don't know how applicable it is here because it has to do with real life data sets, but in short, the leading digits of numbers are distributed in non-uniform ways, with 1 being the most common at around 30% and by the time you get to 9, around 5%. This seems to imply some kind of positive density, which would mean the sum would converge.

Dick
Homework Helper
this is interesting - in my probability theory class today we learned about Benford's law - I don't know how applicable it is here because it has to do with real life data sets, but in short, the leading digits of numbers are distributed in non-uniform ways, with 1 being the most common at around 30% and by the time you get to 9, around 5%. This seems to imply some kind of positive density, which would mean the sum would converge.
You mean since it has positive density it will DIVERGE, right? Benford's law is interesting and all, but it doesn't apply to the list of all numbers. Between 10 and 99, ten numbers start with 1, ten numbers start with 2, ... and 10 numbers start with 9. That's pretty much the whole story.

yeah sorry I meant diverge. My professor has been using the terminology "sparse" and "ubiquitous", so sometimes I lose track

numbers starting with nine have an undefined density, since the limit oscillates

Dick