Date: 04/28/2003 at 08:40:35 From: John Subject: The harmonic series Prove that if in the sum 1 + 1/2 + 1/3 + ... + 1/n we throw out each term that contains 9 as a digit in its denominator, then the sum of the remaining terms is < 80. As n tends to infinity surely the sum will also tend to infinity, as the classic harmonic series does. Perhaps induction on n? The next term will have two possibilities (i.e. either it contains a 9 in the denominator, or it doesn't). In the case where a 9 is present, the inductive hypothesis holds (assuming of course the base case works), but for the latter case I have difficulty showing this to be true.
Date: 04/28/2003 at 11:28:23 From: Doctor Rob Subject: Re: The harmonic series Thanks for writing to Ask Dr. Math, John. Contrary to what you might think, the series you get by deleting all those terms *does* converge, and does *not* approach infinity. This is because as the term numbers increase, the terms remaining comprise a decreasing fraction of all the terms. Partition the series into parts according to how many digits there are in the denominators. The first part is 1/1 + 1/2 + ... + 1/8. It has 8 terms. Each term of this part is less than or equal to 1; all but the first are strictly less. The sum of this part is thus less than 8. (Actually it is even smaller, less than 2.72.) The second part is 1/10 + 1/11 + ... + 1/88. It has 72 = 8*9 terms, because the first digit can be 1 through 8 (8 choices) and the second can be 0 through 8 (9 choices). Each term of this part is less than or equal to 1/10; all but the first are strictly less. The sum of this part is thus less than 8*9*(1/10). The third part is 1/100 + 1/101 + ... + 1/888. It has 8*9^2 = 648 terms, because there are 8 choices for the first digit and 9 choices for both the second and third digits. The terms are each <= 1/100; all but the first are < 1/100. The sum of this part is thus less than 8*9^2*(1/100). In general, the nth part has 8*9^(n-1) terms. They are all <= 1/10^(n-1), and all but the first are <. That means that the sum you seek S satisfies S < 8*1 + 8*9*1/10 + 8*9^2*1/100 + 8*9^3*1/1000 + ... = 8 + 8*(9/10) + 8*(9/10)^2 + 8*(9/10)^3 + ... This is a geometric series with first term 8 and common ratio 9/10. Feel free to write again if I can help further. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
Date: 04/29/2003 at 06:49:15 From: John Subject: The harmonic series Thanks doctor, that's a great help. This is quite an interesting one, though. Is it possible to generalise this to deleting terms containing some other digit? Thanks again.
Date: 04/29/2003 at 09:19:58 From: Doctor Rob Subject: Re: The harmonic series Yes, of course it is possible. Digits 1 through 8 would be exactly the same. Digit 0 would be slightly different, because 0 cannot be the leading digit of any denominator. Feel free to write again if I can help further. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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