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- From: mgh3@po.cwru.edu (mike hurley)
- Subject: Re: Still another problem.
- Message-ID: <mgh3-170892130137@math26647.math.cwru.edu>
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- Organization: case western reserve u
- References: <1992Aug11.170858.275@csc.canterbury.ac.nz> <1992Aug12.075304.28486@newssrv.edvz.univie.ac.at> <1992Aug14.142149.16686@mcs.drexel.edu>
- Date: Mon, 17 Aug 92 17:08:24 GMT
- Lines: 66
-
- In article <1992Aug14.142149.16686@mcs.drexel.edu>, dmagagno@mcs.drexel.edu
- (David Magagnosc) wrote:
- >
- > In article <1992Aug12.075304.28486@newssrv.edvz.univie.ac.at> pm@katz.cc.univie.ac.at (Peter Marksteiner) writes:
- > >Bill Taylor writes:
- > >
- > >
- > >>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.
- > >
- << text deleted >>
- >
- >
- > While we're at it, prove that the following sum is always an
- > integer:
- >
- > n n n n
- > 1 2 3 4
- > -- + -- + -- + -- + ...
- > 1 2 3 4
- > 2 2 2 2
- >
- > and find reasonable generalizations.
- >
- > D. Magagnosc
- > --
-
-
- The new problem can be solved by the same method as the original
- one, namely by expanding using binomial coefficients, changing
- the order of summation, and using induction to obtain a
- recursive formula for the values of the sums:
-
- Let a_n = sum_{j=1}^{infty}(j^n)/(2^j).
-
- Clearly a_0 = 1 is an integer.
- Let B[n,m] denote the binomial coefficient n over m.
-
- We can write
- a_n = (1/2)*\sum_{j=1}^{infty}(j^n)/(2^(j-1))
- = (1/2)*\sum_{j=0}^{infty}((j+1)^n)/(2^j)
- = (1/2)*\sum_{j=0}^{infty}\sum_{m=0}^n B[n,m](j^m)/(2^j)
- = (1/2)*\sum_{m=0}^n B[n,m]\sum_{j=0}^{infty}(j^m)/(2^j).
- For simplicity break off the m=0 term:
- = (1/2)\sum_{j=0}^{infty}(j^0)/(2^j)
- + (1/2)*\sum_{m=1}^n B[n,m]\sum_{j=1}^{infty}(j^m)/(2^j)
- = 1 + (1/2)*\sum_{m=1}^n B[n,m]*a_m.
- Combine both the a_n terms on the left:
- (1/2)a_n = 1 + (1/2)*\sum_{m=1}^(n-1) B[n,m]*a_m
- a_n = 2 + \sum_{m=1}^(n-1) B[n,m]*a_m
- = 1 + \sum_{m=0}^(n-1) B[n,m]*a_m
- (since B[n,0]*a_0 is always 1).
-
- Thus by induction every a_n is an integer; the
- sequence begins 1,2,6,26,150,... .
-
- mike hurley mgh3@po.cwru.edu
- dept of mathematics
- case western reserve univ. 44106-7058
-