superprismatic Posted July 15, 2010 Report Share Posted July 15, 2010 Let an be the nth positive integer which does not contain a 0 digit in its base 10 representation. Does ∞ Σ (1/an) converge or diverge? Why? n=1 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 15, 2010 Report Share Posted July 15, 2010 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) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 15, 2010 Report Share Posted July 15, 2010 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? Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted July 15, 2010 Author Report Share Posted July 15, 2010 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. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 15, 2010 Report Share Posted July 15, 2010 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. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 15, 2010 Report Share Posted July 15, 2010 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 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 15, 2010 Report Share Posted July 15, 2010 (edited) 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 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 15, 2010 Report Share Posted July 15, 2010 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... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 15, 2010 Report Share Posted July 15, 2010 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. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 15, 2010 Report Share Posted July 15, 2010 (edited) 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/an is of the same growth as Σ 1/n. As Σ 1/n diverges, so does Σ 1/an Edited July 15, 2010 by AnyMouse Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 16, 2010 Report Share Posted July 16, 2010 the probability of a number, n, not having a zero in it (could be wrong) = .9log10n 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)*.9log10n this can be "simplified" to E(1/n)*9log10n*10-log10n = E(1/n)*9log10n*10log101/n = E(1/n)*9log10n*1/n = E1/n2*9log10n 9log10n is always less than 1, so by the direct comparison test, E1/n2, which converges, is always greater than E(1/n)*.9log10n, therefore the series converges... maybe Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 16, 2010 Report Share Posted July 16, 2010 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/an is of the same growth as Σ 1/n. As Σ 1/n diverges, so does Σ 1/an Not necessarily.. For every positive integer n, there is an integer n2. But Σ 1/n diverges and Σ 1/n2 converges Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 16, 2010 Report Share Posted July 16, 2010 the probability of a number, n, not having a zero in it (could be wrong) = .9log10n 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)*.9log10n this can be "simplified" to E(1/n)*9log10n*10-log10n = E(1/n)*9log10n*10log101/n = E(1/n)*9log10n*1/n = E1/n2*9log10n 9log10n is always less than 1, so by the direct comparison test, E1/n2, which converges, is always greater than E(1/n)*.9log10n, 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. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 16, 2010 Report Share Posted July 16, 2010 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/an above and below by a constant multiple of 1/n gives you a comparison in BOTH directions. Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted July 17, 2010 Author Report Share Posted July 17, 2010 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! Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted July 19, 2010 Author Report Share Posted July 19, 2010 It converges to something less than 90. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 19, 2010 Report Share Posted July 19, 2010 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...) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 20, 2010 Report Share Posted July 20, 2010 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) Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted July 21, 2010 Author Report Share Posted July 21, 2010 My explanation that the sum converges: There are 9k 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 10k-1. So, the reciprocal of each of these numbers is less than or equal to 1/10k-1. So the contribution of k-digit numbers to our sum is less than or equal to 9k/10k-1. Thus, ∞..............∞ Σ (1/an) ≤ 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. Quote Link to comment Share on other sites More sharing options...
Question
superprismatic
Let an be the nth positive integer which
does not contain a 0 digit in its base
10 representation. Does
∞
Σ (1/an) converge or diverge? Why?
n=1
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