Reciprocals of Integers Greater Than 1 as Sum of a SeriesDate: 07/01/2004 at 13:57:56 From: Nedzad Subject: Expressing the reciprocal of every integer greater than 1 Dear Dr. Math, I have been trying to solve the following problem for the past several hours: Show that the reciprocal of every integer greater than 1 is the sum of a finite number of consecutive terms of the series inf 1 SUM -------- j=1 j(j + 1) I don't even know where to begin... Date: 07/02/2004 at 08:22:24 From: Doctor Jacques Subject: Re: Expressing the reciprocal of every integer greater than 1 Hi Nedzad, Let us write the general term as a(n): a(n) = 1/(n(n + 1)) and the sum of k consecutive terms as s(n,k): s(n,k) = a(n) + ... + a(n + k-1) This kind of series is called a "telescoping sum" and some simplifications happen when you add terms together. We have: s(n,1) = a(n) = 1/(n(n+1)) s(n,2) = a(n) + a(n+1) = 1/(n(n+1)) + 1/((n+1)(n+2)) = (n+2+n) / (n(n+1)(n+2)) = 2(n+1) / (n(n+1)(n+2)) = 2/(n(n+2)) There seems to be a pattern emerging here; maybe it might be true that: s(n,k) = k/(n(n + k)) [1] In fact, this is true, and I would first suggest that you try to prove it. Induction is a good method to try in such a case. The base case (k = 1) is true; assume (induction hypothesis) that: s(n,k) = k/(n(n + k)) and try to prove that the formula is true when you replace k by k + 1, i.e.: s(n,k+1) = (k+1)/(n(n + k+1)) To prove that, use the fact that, by definition: s(n,k+1) = s(n,k) + a(k+1) = s(n,k) + 1/((n+k)(n+k+1)) = k/(n(n+k)) + 1/((n+k)(n+k+1)) Once this is proven, we are reduced to showing that we can find positive integers n and k such that: s(n,k) = 1/a where a > 1 is a given integer. Using [1], this means that we must have: a = n(n+k)/k = (n^2 + nk)/k [2] As this must be an integer, k must divide n^2. The easiest choice that would ensure this would be to try k = n. In this case, we have: a = (k^2 + k^2)/k = 2k If a is even, this gives a solution; take n = k = a/2. However, this does not work if a is odd. The next possible choice to ensure that [2] gives an integer is to take k = n^2. In this case, we have: a = (n^2 + n^3)/n^2 = n + 1 and this works in all cases; take n = a - 1 and k = n^2. For example, with a = 3, we have n = 2 and k = 4: s(2,4) = 1/(2*3) + 1/(3*4) + 1/(4*5) + 1/(5*6) = 1/6 + 1/12 + 1/20 + 1/30 = 1/3 Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/