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- Newsgroups: sci.math
- Path: sparky!uunet!ornl!rsg1.er.usgs.gov!darwin.sura.net!wupost!uwm.edu!linac!uchinews!ellis!glue
- From: glue@ellis.uchicago.edu (brian wesley glueck)
- Subject: Re: 1+1/2+1/3+1/4+...+1/n
- Message-ID: <1992Nov19.163223.23705@midway.uchicago.edu>
- Keywords: partial sums of the harmonic series
- Sender: news@uchinews.uchicago.edu (News System)
- Reply-To: glue@midway.uchicago.edu
- Organization: University of Chicago
- References: <92324.132329K3032E2@ALIJKU11.BITNET>
- Date: Thu, 19 Nov 1992 16:32:23 GMT
- Lines: 22
-
- In article <92324.132329K3032E2@ALIJKU11.BITNET> Mutter Christoph Johannes <K3032E2@ALIJKU11.BITNET> writes:
- >I've a problem. I have to calculate the sum 1+1/2+1/3+1/4+...+1/n.
- >The result should be 100. But my computer is far too slow, to solve this
- >problem. I need the index n when the sum > 100.0
- >And that exactly.
- >Has anyone calculated this? (Perhaps on a CRAY|) ?
-
- Since log(x) is the integral of the reciprocal function from 1 to x, the
- partial sums of the harmonic series can be used to give upper and lower
- bounds for log(x). Draw a picture if you need to in order to see that
-
- 1/2 + 1/3 + ... + 1/n <= log(n) <= 1 + 1/2 + 1/3 + ... + 1/(n-1)
-
- So if log(n) is larger than 100, then the sum on the right is also larger
- than 100, which suggests that you might want to look at values of n that
- are close to exp(100). Of course, exp(100) is a pretty big number (on the
- order to 10^43, I think), but a little analysis helps you get an idea of
- how big n should be without doing any of the summations at all, with or
- without a Cray.
-
- Brian Glueck
- b-glueck@uchicago.edu
-