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Proving Series Convergence


Date: 03/08/98 at 20:42:44
From: Brit Wolfson
Subject: Infinite Series Convergence

In this problem, I am given the infinite series 1/a(n), where a(n) are 
the positive integers that do not contain a 2. We are to show that 
this series converges, and are given one hint to solve the problem: 
break up the sequence of positive integers into blocks, and decide 
what fraction of them is left after removing those containing 2's, and 
then to bound the size of the sum in each block, and try to compare it 
to a different convergent series.

So far, we have decided that the fraction left after removing those 
with 2's is 9^x/10^x, found by taking 10, 100, 1000, and 10000 
positive integers. 

Once we are done with that, we are to find a bound for the sum of the 
series.

Thank you in advance,

Brit Wolfson


Date: 03/09/98 at 13:21:53
From: Doctor Nick
Subject: Re: Infinite Series Convergence

Hi Brit -

You're definitely on the right track with this problem.

One way to bound the sum is the following.

Split the integers into blocks of numbers with the same number of 
digits. So block 1 would be 1,2,3,4,5,6,7,8,9; block 2 would be 
10,11,12,...,99; and so on. Let's consider block d, where d is some 
positive number. How many numbers are in block d? Well, we have 9 
choices for the first digit, and 10 choices for all the other digits. 
So there are:
 
     9*10^(d-1) 

numbers in block d.

How many of these numbers don't have a 2 in their digits? Well, for a 
number in block d, we'd have 8 choices for the first digit, and 9 
choices for all the other digits, so we'd have

     8*9^(d-1)

numbers in block d with no 2 amongst their digits.

Now, the smallest number in block d is 10^(d-1). So the largest 1/n 
can be for n in block d is 1/(10^(d-1)). Therefore, if we add 1/a(n) 
up for all n in block d, we know the sum is less than

    8*9^(d-1) * 1/(10^(d-1)) = 8 * (9/10)^(d-1).

Summing this last expression over d (from 1 to infinity) gives a 
convergent geometric series, whose sum is greater than the sum of 
1/a(n), and so gives a bound for this series.

(I never used the fact that there are 9*10^(d-1) numbers in block d, 
but it's a helpful starting point.)

Have fun,

-Doctor Nick, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
High School Sequences, Series

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