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Nth Term of a Series


Date: 08/27/97 at 07:45:01
From: Alice
Subject: Question!

1/(1*2*3)+1/(2*3*4)+.......+1/<k(k+1)(k+2)>+.......+1/(100*101*102) =?

Thanks!


Date: 08/27/97 at 08:46:40
From: Doctor Anthony
Subject: Re: Question!

                                1
The nth term is of the form  --------
                             n(n+1)(n+2)

                                1          1
We can split this into (1/2)[ -----  - ----------]
                              n(n+1)   (n+1)(n+2)


The terms of the series can be written:

nth term      =  (1/2)[1/n(n+1)     -   1/(n+1)(n+2)]
(n-1)th term  =  (1/2)[1/(n-1)n     -   1/n(n+1) ]
(n-2)th term  =  (1/2)[1/(n-2)(n-1) -  1/(n-1)n ]
------------     ----------------------------------
------------     ----------------------------------
3rd term      =  (1/2)[1/((3)(4))   -   1/((4)(5))]
2nd term      =  (1/2)[1/((2)(3))   -   1/((3)(4))]
1st term      =  (1/2)[1/((1)(2))   -   1/((2)(3))]
-----------------------------------------------------    Add all terms
 Total        =  (1/2)[1/((1)(2)]   -   (1/2)[1/(n+1)(n+2)]


These are the only terms that remain.  All the others cancel between 
rows.

The formula is          SUM = (1/2)[1/2 - 1/(n+1)(n+2)]

If n = 100 this gives   SUM = (1/2)[1/2 - 1/(101)(102)]

                            = 1/4 - 1/20604

                            = 2575/10302      (=.249951465..)

The series converges to the value 1/4 as n -> infinity. 

-Doctor Anthony,  The Math Forum
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Associated Topics:
High School Sequences, Series

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