Sum Term Reciprocals (Posted on 2008-02-22)

	Submitted by K Sengupta
Rating: 3.0000 (2 votes)
Solution:	(Hide)
The required limit is equal to 2. EXPLANATION: We observe that: B(m+1) -1 = B(1)*B(2)* ..*B(m), and: B(m+2) -1 = B(1)*B(2)* ..*B(m+1), and accordingly: B(m+1) = (B(m+2) -1)*(B(m+1) -1)^-1, for m> =2 or, (B(m+2) -1)^-1 = (B(m+1) -1)^-1*B(m+1)^-1 = (B(m+1) -1)^-1 - B(m+1)^-1, for m=2, 3, . Now, B(t+2) = 1 + B(1)*B(2)* *B(t+1) Or, (B(t+2) – 1)^-1 = (B(1))^-1*(B(2))^-1* *(B(t+1))^-1 It can easily be established by induction that each of B(1), B(2), , B(t+2) is >2, and accordingly: Limit (B(t+2) – 1)^-1 t ->∞ = Limit (B(1))^-1*(B(2))^-1* ..*(B(t))^-1    t ->∞ = 0 Also, B(3) = 1 + B(1)*B(2) = 1 + 1*2 = 3 Thus, Limit 1/B(3) + 1/B(4) + ......+ 1/B(t) t-> ∞ = Limit Sum(m = 2 to t) (B(m+1) – 1)^-1 – (B(m+2) – 1)^-1)    t-> ∞ = (B(3) – 1)^-1 Thus, the required limit = (B(3) – 1)^-1 + 1/B(1) + 1/B(2) = (B(3) – 1)^-1 + 1 + 1/2 = 1/2 + 3/2 = 2 Note: This sequence is similar to Sylvester’s Sequence. The difference is that the first term of the given sequence is 1, whereas in case of Sylvester’s Sequence the first term is 2.

	Subject	Author	Date
	Solution (worked out)	FrankM	2008-02-23 07:46:41
	Solution	Dej Mar	2008-02-22 11:49:36
	computer solution -- not proof (spoiler)	Charlie	2008-02-22 11:35:19