Diagonal Sum in Pascal's Triangle

Date: 04/02/2001 at 23:35:43
From: Gene Potter
Subject: Pascal's Triangle

Please explain how to find the sum of the reciprocals of the diagonals 
of Pascal's triangle.

Date: 04/03/2001 at 19:54:43
From: Doctor Schwa
Subject: Re: Pascal's Triangle

Hi Gene,

so you mean, for instance,

     1/1 + 1/3 + 1/6 + 1/10 + ...

or in more compact math language,

     Sum [2/(n(n+1))]

The nifty trick here is to "un-common-denominatorify" this fraction. 
That is, it now has a common denominator of n(n+1). What fractions 
A/n + B/(n+1) would add up to make this answer? (This method is known 
as partial fractions, and you can get help with it in the Dr. Math 
archives by searching for "partial fractions.")

It turns out that 2/n - 2/(n+1) will do the trick.

That is, you can rewrite

     1/1  = 2/1 - 2/2
     1/3  = 2/2 - 2/3
     1/6  = 2/3 - 2/4
     1/10 = 2/4 - 2/5
      :

Now the sum is easy to see: All the terms but 2/1 cancel, so the 
infinite sum is 2/1. The finite sum of the first n terms will be:

     2/1 - 2/(n+1)

What about the next diagonal? You could try a similar trick, but the 
algebra is bound to get messy. Well, not really:

     6/(n(n+1)(n+2)) = 3/n - 6/(n+1) + 3/(n+2)

If you have trouble doing that partial fractions trick, the archives 
will help. Here is a good starting point:

   What is a Partial Fraction?
   http://mathforum.org/dr.math/problems/ishmael1.8.99.html   

So again:

     1/1  = 3/1 - 6/2 + 3/3
     1/4  = 3/2 - 6/3 + 3/4
     1/10 = 3/3 - 6/4 + 3/5
     1/20 = 3/4 - 6/5 + 3/6
      :

By combining the terms with like denominators, you can see that this 
equals 3/1 - 3/2 in the long run; all the rest cancel, and the answer 
is 3/2.

Now if you want a general formula for the k'th diagonal, I'd suggest 
doing one or two more by hand this way (it's not that hard), and then 
looking for a pattern and trying to prove it once you know what the 
answer is already... or write us back. This is an interesting problem. 
If you make some progress but get stuck somewhere along the line, 
please do let me know.

I'm sure the answer is known, and I could just look it up somewhere 
along with a convenient trick for evaluating it, but it's more fun to 
try to figure it out for ourselves.

Thanks for the fun problem,

- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/