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- From: mgh3@po.cwru.edu (mike hurley)
- Subject: Re: Still another problem.
- Message-ID: <mgh3-120892110648@math26647.math.cwru.edu>
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- Organization: case western reserve u
- References: <1992Aug11.170858.275@csc.canterbury.ac.nz>
- Date: Wed, 12 Aug 92 15:10:12 GMT
- Lines: 35
-
- In article <1992Aug11.170858.275@csc.canterbury.ac.nz>,
- wft@math.canterbury.ac.nz (Bill Taylor) wrote:
- >
- > Prove that
- >
- > n n n n
- > 1 / 2 3 4 5 \
- > - | 1 + -- + -- + -- + -- +.... |
- > e \ 1! 2! 3! 4! /
- >
- > is an integer for all positive integer n.
- >
- > ---------------------------------------------------------------------
- > Bill Taylor wft@math.canterbury.ac.nz
- > ---------------------------------------------------------------------
- > Stamp out silly signatures!
- > ---------------------------------------------------------------------
-
-
- Let K(n) denote the indicated quantity;
- (i.e., K(n) = \sum_{j=0}^{infty} (j+1)^{n}/j!/e )
- Then
- K(n+1) = \sum_0^{n+1} B(n+1,m)*K(m-1)
- where B(i,j) is the binomial coefficient i over j
- and K(-1) is defined to be 1. Since K(0)=1 it follows
- that K(n) is integral for every n>=0.
-
- The proof goes by expanding (j+1)^n using the binomial
- theorem, interchanging the order of summation and
- re-indexing.
-
- mike hurley mgh3@po.cwru.edu
- dept of mathematics
- case western reserve univ. 44106-7058
-