1/2 = 1
A proof without words: on one hand,
1/1·3 + 1/3·5 + 1/5·7 + ...  = ½[(1/1  1/3) + (1/3  1/5) + (1/5  1/7) + ...]  
= ½[1/1 + (1/3  1/3) + (1/5  1/5) + ...]  
= ½[1 + 0 + 0 + ...]  
= 1/2. 
On the other hand,
1/1·3 + 1/3·5 + 1/5·7 + ...  = (1/1  2/3) + (2/3  3/5) + (3/5  4/7) + ...  
= 1/1 + (2/3  2/3) + (3/5  2/5) + ...  
= 1 + 0 + 0 + ...  
= 1. 
Therefore, 1/2 = 1!.
Just to make sure that the derivations are absolutely clear, the first one is based on the formula
 = 

The second derivation is based on another formula
 = 

The verification of both formulas is straightforward.
References
 G. P. Vennebush, Math Jokes 4 Math Folks, Robert D. Reed Publishers, 2010, p. 89
Contact Front page Contents Algebra Up
Copyright © 19962018 Alexander Bogomolny
What Went Wrong?
The first derivation is correct, the second is not.
To see why this is so, let's omit the parentheses and evaluate the partial sums.
The first series to consider is
1/1  1/3 + 1/3  1/5 + 1/5  1/7 + ...
(This is twice what we had above.) The partial sums form a convergent sequence: 1, 2/3, 1, 4/5, 1, 6/7, ... The sequence converges to 1. Grouping pairs of successive terms leads to every other sum being omitted but does not change the fact of convergence, nor affects the limit itself.
The second series to consider is
1/1  2/3 + 2/3  3/5 + 3/5  4/7 + ...
The sequence of partial sums is 1, 1/3, 1, 2/5, 1, 3/7, ... does not have a limit. (It does have two accumulation points: 1 and 1/2.) The series therefore is divergent. Grouping pairs of successive terms simply conceals the fact of its divergence. The situation is quite similar to the more obvious conundrum:
1  1 + 1  1 + 1  ... = (1  1) + (1  1) + ... = 1  (1  1)  (1  1)  ...
Contact Front page Contents Algebra Up
Copyright © 19962018 Alexander Bogomolny
64998626 