if the denominator is an even number,n, what is the sum of the series in terms of n

Question

1/4 + 3/4= 1

1/6 +3/6+5/6=3/2

1/8+3/8+5/8+7/8=2

Answer 1

Denote the sum by S.  So 

 

S = 1/n + 3/n + 5/n + ... + (n-5)/n + (n-3)/n + (n-1)/n 

 

Note that there are n/2 terms in the sum.  

 

Multiply both sides by n to get 

 

Sn = 1 + 3 + 5 + ... + (n-5) + (n-3) + (n-1) 

 

The sum on the right is the sum of the first n/2 odd numbers.  If you know that this sum is equal to (n/2)^2 you can write 

 

Sn = (n/2)^2 = n^2/4 which implies that the sum of the original series is S = n/4

 

If you don't know the formula for the sum of the first n/2 odd numbers you can pair up the terms on the right.  The first term and the last term add up to n, as do the 2nd term and the second to last term, and so on.  If n/2 is even then there are n/4 such pairs that add up to n so their sum is (n/4)*n = n^2/4.  If n/2 is odd then there are ((n/2)-1)/2 such pairs plus the middle term which is unpaired.  The middle term is equal to n/2.  So the sum of all the terms is 

 

((n/4) - (1/2)) * n + n/2 = n^2/4 - n/2 + n/2 = n^2/4

 

So in either case Sn = n^2/4  ==>  S = n/4 is the sum of the original series.