Alright Veeky Forums, can you solve this?
s(n) is the digit sum of n.
Alright Veeky Forums, can you solve this?
Julian Adams
Alexander Thompson
Infinity duh.
Jordan Nelson
s(n) = (0,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,10,2,...)
too lazy to prove, but every number k appears exactly k+1 times. If we assume absolute convergence, we can rearrange the sum to be
[math]\sum_{n=0}^{\infty}n(n+1)[/math]
Kayden Long
Shouldn't this diverge (independent of the base used)? s(n) contains all the numbers in N and more, so the sum of the reciprocals of s(n) should be greater than the sum of the reciprocals of N which we know diverges.
In other words (wokring in base 10):
[math]\sum_{n=1}^{\infty} \frac{1}{s(n)} > \frac{1}{1} + \frac{1}{11} + \frac{1}{111} + \frac{1}{1111} + \dots = \sum_{n=1}^{\infty} \frac{1}{n} > \infty
Joseph Thompson
of course its 1/(n(n+1)).
not quite sure what it's evaluated as, but the sum over 1/n^2 is pi^2/6 for example
Tyler Gray
If both of you are correct, then it goes to infinity.
[math]\sum_{n=1}^{\infty}n(n+1)=1[/math]
If n starts from 0, it diverges.
Christopher Robinson
isn't it divergent? The harmonic series is divergent but in this the elements are not getting smaller just repeating themselves.
Jaxon Martin
[math]s(n)\leq n[/math]
hence
[math]\sum_{n=1}^\infty \frac{1}{s(n)}\geq \sum_{n=1}^\infty \frac{1}{n}[/math]
James Long
Consider just the subset of the natural numbers of the form [math] 10^k[/math].
s of this is always 1.
E.g. for [math] 10^5=100000[/math], you have [math]s(10^5)=1+0+0+0+0=1[/math].
So
[math] \sum_{k=0}^\infty \dfrac {1} {s(10^k)} = \sum_{k=0}^\infty 1 = \infty [/math]
Christopher Green
1/1+1/2+1/3+...+1/9+1/1+1/2+1/3+... > 1/1+1/1+... = inf