How to show this series converges?

These problems where you leave terms out of the harmonic series based on the decimal representation are interesting, hard to grasp at first, but easy once you've seen it once.

Let's group the terms by number of digits in the denominator. Considering the numbers as n-letter strings on an alphabet of 9 letters (0 to 8), the group of n-digit numbers without the digit 9 in them has no more than 9^n numbers. And each of them is less than 1/10^(n-1). So our sum is less than the sum over n of

9^n / 10^(n-1)

= 9* sum( (9/10)^n )

which converges.

Oh, one word about the grouping I did. Associativity doesn't always work with infinite series (just consider 1-1+1-1+... ). We can however use associativity when the terms are positive and monotonically decreasing as they are here.

EDIT: Ah, thanks Jake. I was taking the latter sum to be from n=0, but didn't make that clear. Of course, just recognizing it as a geometric sum (with ratio <1) is enough here to claim convergence.

? ? ((-a million)^(n+a million))*(a million/n) = -a million+a million-a million/2+a million/3-a million/4+a million/5-.... n=0 to teach a chain converges we could desire to nicely known the guy words convert to 0, which they do. yet that's not sufficient. ? ? ((-a million)^(n+a million))*(a million/n) = (-a million+a million)+(-a million/2+a million/3)+(-a million/4+a million/5)+ n=0 (-a million/6+a million/7).... =0 -a million/6 -a million/20 -a million/40 two -a million/seventy two-... Factoring the denominators: =0 -a million/(2*3) -a million/(4*5) -a million/(6*7)-a million/(8*9) Noting those components. The words have absolute values smaller than: ? ? (a million/n^2) = a million+a million/4+a million/9+a million/sixteen, which converges n=a million So our sequence could desire to converges. wish this helps. the final analysis is to teach that the sequence has words closer to 0 than a comprehend convergent sequence.

? ? ((-a million)^(n+a million))*(a million/n) = -a million+a million-a million/2+a million/3-a million/4+a million/5-.... n=0 to tutor a series converges we could desire to appreciate the guy words convert to 0, which they do. yet that isn't sufficient. ? ? ((-a million)^(n+a million))*(a million/n) = (-a million+a million)+(-a million/2+a million/3)+(-a million/4+a million/5)+ n=0 (-a million/6+a million/7).... =0 -a million/6 -a million/20 -a million/40 two -a million/seventy two-... Factoring the denominators: =0 -a million/(2*3) -a million/(4*5) -a million/(6*7)-a million/(8*9) Noting those factors. The words have absolute values smaller than: ? ? (a million/n^2) = a million+a million/4+a million/9+a million/sixteen, which converges n=a million So our series could desire to converges. desire this helps. the secret's to tutor that the series has words nearer to 0 than a understand convergent series.

One problem with Ben's solution:

9^n / 10^(n-1)

= 9* sum( (9/10)^n )

SHOULD BE

9^n / 10^(n-1)

= 10* sum( (9/10)^n )

However, the answer remains the same.