Negative powers of 4
Let A = 1 − 4 + 2 − 4 + 3 − 4 + 4 − 4 + 5 − 4 + ⋯ denote the sum of the reciprocals of the fourth powers of all positive integers, and B = 1 − 4 + 3 − 4 + 5 − 4 + 7 − 4 + ⋯ a similar sum for all odd positive integers.
Let B A = D C , where C and D are coprime positive integers, find C + D .
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3 solutions
Thank you for sharing your solution.
no problem
Observe that
A = 1 + 1 6 1 + 8 1 1 + 2 5 6 1 + .....
A = B + 1 6 1 (A)
(A) - 1 6 1 (A) = B
1 6 1 5 A = B
B A = 1 5 1 6
So a+b = 31
N o t e :
This procedure is valid only if the sum converges
True, thank you for the nice solution.
Also add a note saying that this willl be valid only if the sum converges.
1 6 / 1 5 = A / B ⟹ A + B = 3 1
B = A − ( 2 − 4 + 4 − 4 + 6 − 4 + ⋅ ⋅ ⋅ ) = A − 1 6 A
Only a brief idea that might be helpful
∑ 1 ∞ n − 4 = 9 0 π 4
I have proved it here- https://brilliant.org/problems/the-basel-problem-ii/
using that A = 9 0 π 4
B = A - 2 4 1 ( A )
Now dividing both of them gives:
B A = 1 5 1 6
Guess the infinite sum is not required here but it can still be used to prove that the sum converges