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evonnova
Average Member

USA
13 Posts

Posted - 10/11/2007 :  20:05:26  Show Profile
Let S={n, n,.....} denote the collection of positive integers that do not involve the digit 0 in their decimal expansion. Prove that {k=1 to infinity} 1/n[sub-k] converges and has sum less than 90.
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Ultraglide
Advanced Member

Canada
299 Posts

Posted - 10/11/2007 :  23:00:52  Show Profile
What do you mean by not involving zero in the decimal expansion?
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evonnova
Average Member

USA
13 Posts

Posted - 10/11/2007 :  23:17:38  Show Profile
i really don't know either because my teacher give this problem to solve and he's out of town not going to teach class for a month so..i don't know who i'm going to ask
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someguy
Advanced Member

Canada
143 Posts

Posted - 10/12/2007 :  03:14:42  Show Profile
By having no zero in their decimal expansion, your instructor means that the (finite) base 10 expansion of the integer must not have a zero in it.

For instance, the positive integers 1, 14, 5927113, and 9987718823 are all members of S, but 10, 123012, and 1230123400012348903 are not members of S.

To answer this question, group the numbers in S by the number of digits in their base 10 expansion.

You will see that there are 9 single digit numbers in S.
1,2,3,4,5,6,7,8,9

There are 81 (=9^2) two digit numbers in S.
11,12,13,14,15,16,17,18,19,
21, . . .
31, . . .
41, . . .
51, . . .
61, . . .
71, . . .
81, . . .
91,92,93,94,95,96,97,98,99

There are 9^3 three digit numbers in S.
(Any 3 digit number in S can be formed by placing a 1,2,3,4,5,6,7,8, or 9 in front of one of the allowable 2 digit numbers and any number formed in such a way is in S).

Hopefully you can see that there are 9^n n digit numbers in S.

Now ask yourself how big 1/m can be if m is an n digit positive integer.

The problem reduces down to finding the sum of a friendly series you should be familiar with.
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