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# Harmonic - 14 Â a Now ln N 1< âˆ‘ N n =1< ln N 1 1 For...

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Calculating the Harmonic Series James Keesling The HarmonicSeries is given by the following formula. X n =1 1 n It is well-known that n =1 1 n = . However, if we were to add the terms together, adding the next term to the previous partial sum, the resulting partial sums would not grow to inﬁnity as predicted by pure theory. It is simple to understand why this is so. If a is number represented as a ﬂoating point number in the computer, then there is a number ± ( a ) > 0 such that if 0 < b < ± ( a ), then in the computer a + b = a . So, as the partial sums, N n =1 1 n are approximated by ﬂoating point numbers S N , there is a value of N such that 1 N +1 < ± ( S N ). For that value of N , the partial sums will stop growing and that will be the terminal value of the series as far as the computer is concerned. At what value would this happen on your TI-89? If a is a ﬂoating point number in the TI-89, let us assume that ± ( a ) 10 -
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Unformatted text preview: 14 Â· a . Now, ln( N + 1) < âˆ‘ N n =1 < ln( N + 1) + 1. For N = 10 13 we get ln(10 13 ) â‰ˆ 29 . 933606 ... and 10-14 Â· log(10 13 + 1) â‰ˆ 2 . 9933606 Â· 10-13 > 10-13 . So, the partial sums will stop growing at about N = 10 13 and the terminal value of the sum will be about 30. How long would it take to do this mistaken calculation on your TI-89? A state of the art supercomputer can perform about 10 16 ï¬‚oating-point operations per second. If such a computer carried an accuracy of 10-30 digits, the terminal sum would be about ln(10 3 0) â‰ˆ 69 . 07755 ... and 10-30 Â· ln(10 30 ) â‰ˆ 6 . 907755 Â·Â·Â· < 10-30 . So, the sum would terminate at about 69 . 07755 ... . It would take approximately 10 30 operations. This would take approximately 10 14 seconds or about 3,170,979 years to reach this wrong answer. 1...
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