From: ngoclu@triton.unm.edu (ngoc-dung t  luntanondana)
Newsgroups: soc.culture.vietnamese
Subject: Re:Solution to a math problem in series
Summary: Another way to calculate the sum
Date: 7 Nov 92 21:25:18 GMT
Distribution: usa
Organization: University of New Mexico, Albuquerque

>From lynx!news.cs.indiana.edu!usenet.ucs.indiana.edu!ux1.cso.uiuc.edu!sdd.hp.com!decwrl!netsys!agate!dog.ee.lbl.gov!news!manta!dam Fri Nov  6 13:02:03 MST 1992
>Article: 21275 of soc.culture.vietnamese
>Newsgroups: soc.culture.vietnamese
>Path: lynx!news.cs.indiana.edu!usenet.ucs.indiana.edu!ux1.cso.uiuc.edu!sdd.hp.com!decwrl!netsys!agate!dog.ee.lbl.gov!news!manta!dam
>From: dam@nosc.mil (Khanh Dam)
>Subject: Re: (none)
>Message-ID: <1992Nov4.183520.3088@nosc.mil>
>Summary: just a try
>Organization: Naval Ocean Systems Center, San Diego, CA
>References: <9211040511.AA193825@nike.calpoly.edu>
>Date: Wed, 4 Nov 1992 18:35:20 GMT
>Lines: 51
>
>In article <9211040511.AA193825@nike.calpoly.edu>, tltran@nike.calpoly.edu (ta^m) writes:
>> 
>>       To^i co' ba`i toa'n na`y la`m hoa`i kho^ng ddu*o*.c. Ne^'u co' anh
>> chi. na`o bie^'t la`m o*n help me.(especially part b)
>>   -----
>> 
>>       Given T(1) = 2,
>>               T(n+1) = [T(n)]^2 - T(n) + 1,    n > 0.
>>       Prove:
>> 
>>     a,  If m != n, T(m) and T(n) have no common factor greater than 1
>>     b,  calculate
>>              1/T(1) + 1/T(2) + 1/T(3) + ...
>> 
>>     note:
>>         T(n) mean T subscript n
>>         m != n mean m not equal to n
>> 
>>                                                        Thankx
>>                                                         Ta^m
>> 
>
>Ho^m nay ra?nh ro^?i, -do.c ba`i na`y tha^'y hay hay ne^n Kh. la`m thu+? nhe':

>T(n+1) = T(n)[ T(n) - 1 ] +1,   n > 0
 
>b).
>	 T(1) = 2
>n = 1 => T(2) = T(1)[ T(1)-1 ] + 1
>	      = 2[1] + 1 = 3
>
>n = 2 => T(3) = 3[2] + 1 = 7
>n = 3 => T(4) = 7[6] + 1 = 43
>n = 4 => T(5) = 43[42] + 1 = 1807
>n = 5 => T(6) = 1807[1806] + 1 = 326,3443 
>.
>.
>.
>  => 1/T(1) + 1/T(2) + 1/T(3) + ... = 1/2 + 1/3 + 1/7 + 1/43 + 1/1807 + ....
>				    = 1.0000

>a).
>     for any m != n => T(m) != T(n) (or T(m+1) != T(n+1)) and the terms
>		       {T(m)[ T(m)-1 ] +1}  and  {T(n)[ T(n)-1 ] +1}
>		       will have no common factor > 1 (because of the +1 term).
>
>     this can be shown by induction of the above terms for n = 1, 2, 3, ...
>
>
>C. Kh.
    I have a comment: Ba.n Khanh  tuy ra(`ng ti`m ddu'ng dda'p so^' 
cu?a to^?ng song ca'ch tri`nh ba`y va` ti'nh nhu' va^.y co' ve? e'p qua'.
To^i co' ca'ch na`y, kho^ng bie^'t Ta^m va` Khanh xem co' ddu'o'.c kho^ng
( To+' mo+'i vu+`a ddo.c ba`i toa'n na`y tho^i)
   b-
    let's call S(n)=1/T(1)+...+1/T(n)
    We have
     T(n+1)=T(n)**2-T(n)+1=T(n)[T(n)-1]+1
  => T(n+1)-1=T(n)[T(n)-1]
  Hence,
     T(2)=T(1)[T(1)-1]+1 => T(2)-1=T(1)[T(1)-1]
     T(3)=T(2)[T(2)-1]+1
         =T(2).T(1).[T(1)-1]+1
     .
     .
     .
     T(n-1)=T(n-2)...T(1)[T(1)-1]+1
     T(n)  =T(n-1).T(n-2)...T(1)[T(1)-1]+1
  From this we have:

     T(n)-T(n-1)...T(1)(T(1)-1)= 1                   (*)

     Let's call A(k) an element of the series S(n); then A(k)=1/T(k).

     We set up another series P(n) = sum (from 1 to n) of B(k)
with B(k)=1/[T(k-1).T(k-2)...T(1).(T(1)-1)] for k>1, and
B(1)=1/[T(1)-1] ( for obviously T(1)#1).
( notice that there is no extra 1 in the denominator of B(k)).

     Then, for any k > 1 we have:
     B(k)-A(k)=1/[T(k-1)...(T(1)-1)] - 1/T(k)   
              =[T(k)-T(k-1)...(T(1)-1)] / [T(k)...T(1).(T(1)-1)]
              = 1/[T(k)...T(1).(T(1)-1)]   ( because of (*))
              = B(k+1) ( because of the way we set up B(k))
     Therefore:
     B(1) - A(1) = 1/[T(1)-1]-1/T(1)=B(2)
     B(2) - A(2) = B(3)
     .
     .
     .
     B(n) - A(n) = B(n+1)
  _______________________________________
=>   B(1) - S(n) = B(n+1)
=>   S(n) = B(1)-B(n+1)
     It is easy to see that as  n goes to infinity, B(n+1) goes to 0.
     Therefore,
     S(n) = 1/T(1)+...+ 1/T(n) goes to 1 as n goes to infinity
      or
1/T(1)+ 1/T(2)+........=1