We can list all the fractions in an infinite grid:
1
22
1
32
1
42

1
23
1
33
1
43

1
24
1
34
1
44

1

25
1

35
1

45

Notice that the entries in each row form an infinite geometric progression.
1
2
1
1
1
1
1
1
The fractions in the 1st row sum to 2  3  4  5  ...  2  2

1 2 2 2
2 2 2 2
1
2
1 1 1
1
Similarly:
 3  4  ..  ,
2
3 3 3
6
1 1 1
1
 3  4  .. 
2
4 4 4
12
1 1 1
1
and, in general, 2  3  4  .. 
.
r
r
r
r  r  1
Therefore, the sum of all the fractions in the grid is the sum of the row totals:
1 1 1
1
   ...... 
 ...
2 6 12
r  r  1

So we need to find
1
 r  r  1 .
r 2
The best way to tackle this is to use partial fractions:
1
1
1


r  r  1 r  1 r

and so

1
1
 1

 



r
r  2 r  r  1
r 2  r  1
1 1   1 1   1 1   1 1 
                 ...
1 2   2 3   3 4   4 5 
1
A surprising result!
© MEI 2006