Flip Flop
n = 1 ∑ ∞ n 2 f ( n )
Let f be an injective function maps from the set of positive integers to itself. Choose the correct answer for the value of the series above.
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3 solutions
In applying the rearrangement inequality in the above solution, I think, it has been implicitly assumed that, for every N , f is an injective mapping of the form f : { 1 , 2 , … , N } → { 1 , 2 , … , N } , which restricts an injective f only to permutations. Since f is a permutation for every N , this in turn implies that f : N → N is the identity function. Am I missing something here ?
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Yes, you are missing something. To create a reverse sum, I'm rearranging f ( 1 ) , . . . , f ( N ) in ascending order, f ( σ ( 1 ) ) < . . < f ( σ ( N ) ) , where σ is a permutation of {1,..., N }. Note that f ( σ ( n ) ) ≥ n since f ( σ ( n ) ) is strictly increasing. Now we have ∑ n = 1 N n 2 f ( n ) ≥ ∑ n = 1 N n 2 f ( σ ( n ) ) ≥ ∑ n = 1 N n 2 n , as claimed.
let's suppose that f ( n ) = 2 n , we clearly see that f is injective, we therefore have :
n = 1 ∑ ∞ n 2 f ( n ) = n = 1 ∑ ∞ n 2 2 n = n = 1 ∑ ∞ n 2
which doesn't converge, so the answer has to be "not convergent" or "undetermined" , since there can be only one solution then the answer is "not convergent".
But can you prove this for all f ?
The function n 3 is injective in the positive integers. We quickly see that the sum n = 1 ∑ ∞ n does not converge so the general case is disproved by simple counter example.
I did it with f(n) = n.
Great problem!
By the rearrangement inequality , "random sum ≥ reverse sum", we have ∑ n = 1 N n 2 f ( n ) ≥ ∑ n = 1 N n 2 n = ∑ n = 1 N n 1 , so that the partial sum diverges.