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- Newsgroups: sci.math
- Path: sparky!uunet!newsgate.watson.ibm.com!yktnews!admin!yktnews!victor
- From: victor@watson.ibm.com (Victor Miller)
- Subject: Re: need proof: (1 + 1/n)^n ==> e
- Sender: news@watson.ibm.com (NNTP News Poster)
- Message-ID: <VICTOR.92Dec30104755@terse.watson.ibm.com>
- In-Reply-To: pratt@Sunburn.Stanford.EDU's message of Tue, 29 Dec 1992 09:48:42 GMT
- Date: Wed, 30 Dec 1992 15:47:55 GMT
- Lines: 34
- Reply-To: victor@watson.ibm.com
- Disclaimer: This posting represents the poster's views, not necessarily those of IBM
- References: <1glr2qINN2vg@usenet.INS.CWRU.Edu> <1hfv4sINNdlk@usenet.INS.CWRU.Edu>
- <1992Dec29.094842.3685@CSD-NewsHost.Stanford.EDU>
- Nntp-Posting-Host: terse.watson.ibm.com
- Organization: IBM, T.J. Watson Research Center
-
- The following simple proof comes from "Inequalities" by P. Korovkin,
- Blaidell Scientific paperbacks.
-
- set x_n = (1+1/n)^n, and y_n = (1+1/n)^{n+1}.
-
- We show that x_n is monotone increasing and y_n is monotone
- descreasing. Since x_n < y_n we find that they both approach a common
- limit, which we call e.
-
- Proofs:
-
- We use the inequality of the arithmetic and geometric mean, in the
- following form:
-
- (a b^n)^{1/(n+1)} <= (a + n b)/(n+1).
-
- If we let a=1, and b = (1+1/n), and raise both side to the (n+1)st
- power, we find:
-
- x_n <= ((1 + n(1+1/n))/(n+1))^{n+1} = (1+1/(n+1))^{n+1} = x_{n+1}.
-
- Now set z_n = (1-1/n)^n. As above, we set a=1, and b=(1-1/n), we get
-
- z_n <= ((1 + n(1-1/n))/(n+1))^{n+1} = (1 - 1/(n+1))^{n+1} = z_{n+1}.
-
- Now observe that y_n = 1/z_{n+1}, so y_n is decreasing. QED
-
-
-
- --
- Victor S. Miller
- Vnet and Bitnet: VICTOR at WATSON
- Internet: victor@watson.ibm.com
- IBM, TJ Watson Research Center
-