Infinite series (easy)

Σ(n = 1000 to ∞) 1/(n² - 1).
=(1/2)Σ(n = 1000 to ∞) [1/(n - 1) - 1/(n+1) ].
=(1/2)( 1/999 + 1/1000 )

is rational.


2012-05-24 17:36:09 補充：
To: 002
Σ(n = 1000 to ∞) 2/(n² - 1) is convergent.
Σ(n = 1000 to ∞) 1/(n-1) , Σ(n = 1000 to ∞) 1/(n+1) are divergent
Σ(n=1000 to ∞) 2/(n² - 1) does not equal Σ(n =1000 to ∞) 1/(n-1) - Σ(n=1000 to ∞) 1/(n+1)

2012-05-24 17:41:02 補充：
Detail proof:
Suppose N > 1000
Σ(n = 1000 to N) 1/(n² - 1)
=(1/2) Σ(n =1000 to N) [1/(n-1) - 1/(n+1) ]
=(1/2) [ 1/999+ 1/1000 - 1/N - 1/(N+1) ]
lim(N->∞) (1/2)[ 1/999+ 1/1000 - 1/N - 1/(N+1) ] =(1/2) (1/999+ 1/1000)
so Σ(n = 1000 to ∞) 1/(n² - 1) = (1/2)( 1/999+ 1/1000) is rational.

2/(n² - 1)=1/(n-1)-1/(n+1)

∴ Σ(n = 1000 to ∞) 2/(n² - 1)= Σ(n = 1000 to ∞) 1/(n-1) - Σ(n = 1000 to ∞) 1/(n+1)
=(1/999+1/1000+1/1001+1/1002+...)-(1/1001+1/1002+1/1003+1/1004+...)
=1/999+1/1000+[(1/1001+1/1002+...)-(1/1001+1/1002+1/1003+1/1004+...)]
= 1/999+1/1000+0
= 1/999+1/1000
=1999/999000
∴ Σ(n = 1000 to ∞) 1/(n² - 1)= (1999/999000)/2=1999/1998000

It is an fractional number
∴ it is irrational (definition of rational number)