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 1014 Â· log(10 13 + 1) â‰ˆ 2 . 9933606 Â· 1013 > 1013 . 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 TI89? A state of the art supercomputer can perform about 10 16 ï¬‚oatingpoint operations per second. If such a computer carried an accuracy of 1030 digits, the terminal sum would be about ln(10 3 0) â‰ˆ 69 . 07755 ... and 1030 Â· ln(10 30 ) â‰ˆ 6 . 907755 Â·Â·Â· < 1030 . 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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 Summer '08
 Kyung
 Statistics, Summation, partial sums, 3,170,979 years, 30 digits, 1014 seconds

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