home *** CD-ROM | disk | FTP | other *** search
- Newsgroups: sci.math
- Path: sparky!uunet!zaphod.mps.ohio-state.edu!sdd.hp.com!spool.mu.edu!agate!linus!linus.mitre.org!gauss!bs
- From: bs@gauss.mitre.org (Robert D. Silverman)
- Subject: Re: 1+1/2+1/3+1/4+...+1/n
- Message-ID: <1992Nov21.125218.23926@linus.mitre.org>
- Sender: news@linus.mitre.org (News Service)
- Nntp-Posting-Host: gauss.mitre.org
- Organization: Research Computer Facility, MITRE Corporation, Bedford, MA
- References: <92324.132329K3032E2@ALIJKU11.BITNET> <1992Nov19.133451.11346@hubcap.clemson.edu> <michaelg.722193567@Xenon.Stanford.EDU>
- Date: Sat, 21 Nov 1992 12:52:18 GMT
- Lines: 24
-
- In article <michaelg.722193567@Xenon.Stanford.EDU> michaelg@Xenon.Stanford.EDU (Michael Greenwald) writes:
-
- :(d) Also, since he's looking for the particular "n" that will allow
- :the harmonic series to sum to 100, it will be difficult for him to
- :proceed from "the smallest to the highest" without having a guess as
- :to what "n" is (although one would assume that someone who had taken
-
- Since no partial sum of the harmonic series is ever an integer,
- it is going to be somewhat difficult to find such an 'n'.
-
- Hint for proof: Let H = 1 + 1/2 + .... 1/n, multiply both sides by
- n! and count powers of 2 dividing both sides [or use Bertrand's hypothesis,
- i.e. look at the largest prime dividing both sides]
-
- As for the original question; one can get some very good approximation
- to H = 1 + 1/2 + .... 1/n via Euler-MacLauren summation, but the error
- term isn't good enough to determine the EXACT n that lets the sum first
- exceed 100. You would also need Euler's constant to at least 45 decimal
- places.
- --
- Bob Silverman
- These are my opinions and not MITRE's.
- Mitre Corporation, Bedford, MA 01730
- "You can lead a horse's ass to knowledge, but you can't make him think"
-