Associated Topics || Dr. Math Home || Search Dr. Math

### Reciprocals of Integers Greater Than 1 as Sum of a Series

```Date: 07/01/2004 at 13:57:56
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

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

more, or if you have any other questions.

- Doctor Jacques, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory
High School Sequences, Series

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search