Infinite sums

Date: 6/26/96 at 2:23:26
From: Anonymous
Subject: Toughie

Given the function f(x)= x^2/(1+x^2) find the sum

               f(1/1)+f(2/1)+...+f(n-1/1)+f(n/1)
               f(1/2)+f(2/2)+...+f(n-2/2)+f(n/2)
               f(1/n)+f(2/n)+...+f(n-1/n)+f(n/n)

for any positive integers "N".

Date: 6/26/96 at 11:47:1
From: Doctor Anthony
Subject: Re: Toughie

I assume from this that you require the sum of ALL the terms made up 
of n rows and n columns.

Note first that f(x) + f(1/x) = x^2/(1+x^2) + (1/x^2)/(1+(1/x^2))
                              = x^2/(1+x^2) + 1/(1+x^2)
                              = (x^2+1)/(1+x^2)
                              = 1.

If we consider the terms laid out in n rows and n columns, then the 
leading diagonal has terms like f(1/1), f(2/2)....f(n/n), each equal 
to f(1) = 1/2.

So the terms in the leading diagonal add up to n/2.  Now having 
removed the leading diagonal, you are left with n^2-n terms.  These 
can be paired off, with each pair contributimg 1 to the total.  So 
other terms contribute (n^2-n)/2.  The overall total is therefore 
(n^2-n)/2 + n/2 = n^2/2.

For n even we have the very simple result for n^2/2.

Thus if n = 2 total is 4/2 = 2
     if n = 4 total is 16/2 = 8
     if n = 6 total is 36/2 = 18   and so on.

n ODD.
     if n = 1 total is 1^2/2 = 1/2
        n = 3 total is 9/2 = 4 +(1/2)
        n = 5 total is 25/2 = 12 +(1/2).

With n ODD we shall always have a half as part of the sum.
 
-Doctor Anthony,  The Math Forum
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