Find ∑1∞∑1∞ 1/(i2j + 2ij + ij2).
Solution
Answer: 7/4.
Let us fix i and sum over j. The term is 1/(i(i + 2)) (1/j - 1/(j + i + 2) ). So, summing over j, all the terms cancel except for the first i + 2, giving: 1/(i(i + 2)) (1/1 + 1/2 + ... + 1/(i + 2) ). Also 1/(i(i + 2)) = 1/2 (1/i - 1/(i + 2) ).
So summing over i (and multiplying by 2) we get:
(1/1 - 1/3) (1 + 1/2 + 1/3) +
(1/2 - 1/4) (1 + 1/2 + 1/3 + 1/4) +
(1/3 - 1/5) (1 + 1/2 + 1/3 + 1/4 + 1/5) +
(1/4 - 1/6) (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6) + ...
Each of the minus terms partially cancels with the corresponding plus term two lines lower, so we get:
1/1(1 + 1/2 + 1/3) + 1/2(1 + 1/2 + 1/3 + 1/4) + 1/3(1/4 + 1/5) + 1/4(1/5 + 1/6) + ...
= 1 + 1/4 + 1/2 + (1/1 1/2 + 1/2 1/3 + 1/3 1/4 + ... ) + (1/1 1/3 + 1/2 1/4 + 1/3 1/5 + ... )
= 7/4 + ( (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4 ) + ... ) + 1/2 ( (1 - 1/3) + (1/2 - 1/4) + (1/3 - 1/5) + ... )
= 7/4 + 1 + 1/2 (1 + 1/2) = 7/2.
Hence the answer is 7/4.
© John Scholes
jscholes@kalva.demon.co.uk
30 Nov 1999