[superprismatic]

Let a_n be the n^th positive integer which

does not contain a 0 digit in its base

10 representation. Does

∞

Σ (1/a_n) converge or diverge? Why?

n=1

[Guest]

I think it diverges:

I'm not sure of my thinking, but here goes:

Let's say X = Σ (1/an)

We could also say that X = Y - Z where

Y = Σ (1/n) (n = integers from 1 to ∞)

Z = Σ (1/m) (m = integers that contain 0s)

if Y diverges from Z (i.e. much greater), then it must diverge.

for a m digit number, there are

i

Σ (10^(i-1)) occurrences of numbers with a 0 in them (i.e. a 5 digit number has 11111 numbers with 0 in it).

n=1

5 digit example:

Y contains 88888 values with zeros

Z contains 11111 values with zeros

For 5 digits Y = Σ(1/n) is MUCH greater than Z = Σ(1/m)

Therefor I think it diverges (again, I'm not sure about the thinking)

[Guest]

so to clarify, one example would be...

base 1 would be 1, base 2 would be 2, but then starting at base 10 it gets tricky?

[superprismatic]

so to clarify, one example would be...

base 1 would be 1, base 2 would be 2, but then starting at base 10 it gets tricky?

I'm sorry but I don't understand your question.

[Guest]

I think I'm not familiar with this converge or diverge thing, or maybe I just forgot it. The sum converges if it has a finite limit? In that case, I think this one converges, as it would be kinda similar to the sum of 1/x². However I have no idea what the limit would be, nor am I sure everything I said so far is correct, so I'm gonna stop right now.

[Guest]

I think I'm not familiar with this converge or diverge thing, or maybe I just forgot it. The sum converges if it has a finite limit? In that case, I think this one converges, as it would be kinda similar to the sum of 1/x². However I have no idea what the limit would be, nor am I sure everything I said so far is correct, so I'm gonna stop right now.

Yes... you're right with your definition of converge.

Σ (1/n^2) (n = 1 to inf) converges on pi^2/6

however Σ (1/n) (n = 1 to inf) actually diverges

[Guest]

I wrote some C++ and I've decided that it probably converges to 10...

/*This file is made for brainden.com. The puzzle is does this diverge or converge: the sum from n=1 to inf of 1/An where A is the second number in base 10 representation to not contain a 0 digit*/

#include <iostream>

#include <iomanip>

#include <cmath>

using namespace std;

bool hasnozeros(int );

int main()

{

int a = 0;

int base = 1;

double sum = 0;

int count;

for( base = 0; base < 30000; ++base)

{

count = 0;

for( a = 1; count < base; ++a)

{

if( hasnozeros(a))

++count;

}

sum += (1 / ((double) a));

}

cout << endl << "For n = 1 to 30,000, the answer is: " << sum << endl;

return 0;

}

bool hasnozeros(int x)

{

int temp = x;

while(temp > 10)

{

if( temp % 10 == 0 )

return false;

else

temp /= 10;

}

return true;

}

Edited July 15, 2010 by madscout

[Guest]

I wrote some C++ and I've decided that it probably converges to 10...

/*This file is made for brainden.com. The puzzle is does this diverge or converge: the sum from n=1 to inf of 1/An where A is the second number in base 10 representation to not contain a 0 digit*/

#include <iostream>

#include <iomanip>

#include <cmath>

using namespace std;

bool hasnozeros(int );

int main()

{

int a = 0;

int base = 1;

double sum = 0;

int count;

for( base = 0; base < 30000; ++base)

{

count = 0;

for( a = 1; count < base; ++a)

{

if( hasnozeros(a))

++count;

}

sum += (1 / ((double) a));

}

cout << endl << "For n = 1 to 30,000, the answer is: " << sum << endl;

return 0;

}

bool hasnozeros(int x)

{

int temp = x;

while(temp > 10)

{

if( temp % 10 == 0 )

return false;

else

temp /= 10;

}

return true;

}

Nice program.

If you increase the sum to something much greater what the result is...

You're program looks pretty efficient.

Σ (1/n) (n=1 to 30000) = 10.886

Σ (1/n) (n=1 to 300000) = 13.189

Σ (1/n) (n=1 to 3000000) = 15.491

Σ (1/n) does diverge...

[Guest]

Nice program.

If you increase the sum to something much greater what the result is...

You're program looks pretty efficient.

Σ (1/n) (n=1 to 30000) = 10.886

Σ (1/n) (n=1 to 300000) = 13.189

Σ (1/n) (n=1 to 3000000) = 15.491

Σ (1/n) does diverge...

Thank you for carrying that out, I ran that on my school's UNIX server so when I went to try up to 100,000, it wasn't worth the time anymore.

[Guest]

For each positive integer k, there is a positive integer between k and 2k without a zero in its expansion. This means that

Σ 1/a_n is of the same growth as Σ 1/n.

As Σ 1/n diverges, so does Σ 1/a_n

Edited July 15, 2010 by AnyMouse

[Guest]

the probability of a number, n, not having a zero in it (could be wrong) = .9^log₁₀n and obviously as n increases the chances there are no zeros decreases.

next i lazily assumed that i could represent the sum as follows (which i might not be able to): E(1/n)*.9^log₁₀n

this can be "simplified" to E(1/n)*9^log₁₀n*10^-log₁₀n = E(1/n)*9^log₁₀n*10^log₁₀1/n = E(1/n)*9^log₁₀n*1/n = E1/n²*9^log₁₀n

9^log₁₀n is always less than 1, so by the direct comparison test, E1/n², which converges, is always greater than E(1/n)*.9^log₁₀n, therefore the series converges... maybe

[Guest]

For each positive integer k, there is a positive integer between k and 2k without a zero in its expansion. This means that

Σ 1/a_n is of the same growth as Σ 1/n.

As Σ 1/n diverges, so does Σ 1/a_n

Not necessarily..

For every positive integer n, there is an integer n². But Σ 1/n diverges and Σ 1/n² converges

[Guest]

the probability of a number, n, not having a zero in it (could be wrong) = .9^log₁₀n and obviously as n increases the chances there are no zeros decreases.

next i lazily assumed that i could represent the sum as follows (which i might not be able to): E(1/n)*.9^log₁₀n

this can be "simplified" to E(1/n)*9^log₁₀n*10^-log₁₀n = E(1/n)*9^log₁₀n*10^log₁₀1/n = E(1/n)*9^log₁₀n*1/n = E1/n²*9^log₁₀n

9^log₁₀n is always less than 1, so by the direct comparison test, E1/n², which converges, is always greater than E(1/n)*.9^log₁₀n, therefore the series converges... maybe

I just tried out my approximation and i found that after 999 terms, it is .3 lower than the actual value of the series, which does not allow me to claim convergence.

[Guest]

Not necessarily..

For every positive integer n, there is an integer n². But Σ 1/n diverges and Σ 1/n² converges

because for 1/n and 1/n² you only have a one-sided comparison 1/n² < 1/n.

That you can bound 1/a_n above and below by a constant multiple of 1/n gives you a comparison in BOTH directions.

[superprismatic]

The sum actually converges. Now, can someone prove that? By the way, although I can prove convergence, I have no idea what it converges to!

[superprismatic]

It converges to something less than 90.

[Guest]

I would guess it converges.

heres why: as numbers begin to have more and more digits, the probability that they will have a zero in them approaches one. in fact, the probability is 1-(9/10)^n where n is the number of digits.

if i had to guess what it converged to, i would guess about 80. (I'm not going to give my reasoning for this because it is so convoluted that it would be rather embarrassing...)

[Guest]

I would guess it converges.

heres why: as numbers begin to have more and more digits, the probability that they will have a zero in them approaches one. in fact, the probability is 1-(9/10)^n where n is the number of digits.

if i had to guess what it converged to, i would guess about 80. (I'm not going to give my reasoning for this because it is so convoluted that it would be rather embarrassing...)

the number of digits in an integer = floor(log₁₀n), but i have no idea how to incorporate a floor function into an infinite sum... unless it is mathematically sound to say x-1>floor(x)

[superprismatic]

My explanation that the sum converges:

There are 9^k positive integers not containing

a zero which are k digits long (for each of the

k positions, we can use any of the nine digits

from 1 to 9). Each one of these numbers is

greater than or equal to 10^k-1. So, the

reciprocal of each of these numbers is less

than or equal to 1/10^k-1. So the

contribution of k-digit numbers to our sum is

less than or equal to 9^k/10^k-1. Thus,

∞..............∞

Σ (1/a_n) ≤ Σ 9×(9/10)^k-1 = 90

n=1.........k=1

Note: I'm using periods in the last formula

above only to keep things aligned -- they

have no meaning. Just consider them to be

white space.