1, -1, 2, -4, 3, -9, 4, -16, 5, -25, ... — in general a(2k-1) = k, a(2k) = -k².
Square the odd-indexed terms only: they become 1, 4, 9, 16, 25, ..., and the even-indexed terms are exactly -1, -4, -9, -16, -25, .... Every consecutive pair (k, -k²) contributes k² - k² = 0, so every partial sum is 0 and the series converges trivially to exactly 0.
How I found it: the signature is "square every other term", so I forced each even term to be the negative square of the preceding odd term — cancellation is then termwise, not asymptotic, so no convergence argument is needed. Choosing the odd terms to be the naturals keeps it integral and non-trivial.
Generalization: for any sequence b₁, b₂, b₃, ..., the sequence b₁, -b₁², b₂, -b₂², b₃, -b₃², ... satisfies the property with sum 0. Rational family: b_k = 1/k gives 1, -1, 1/2, -1/4, 1/3, -1/9, ....
Solution (integer sequence, exact zero):
1, -1, 2, -4, 3, -9, 4, -16, 5, -25, ...— in generala(2k-1) = k,a(2k) = -k².Square the odd-indexed terms only: they become
1, 4, 9, 16, 25, ..., and the even-indexed terms are exactly-1, -4, -9, -16, -25, .... Every consecutive pair(k, -k²)contributesk² - k² = 0, so every partial sum is 0 and the series converges trivially to exactly 0.How I found it: the signature is "square every other term", so I forced each even term to be the negative square of the preceding odd term — cancellation is then termwise, not asymptotic, so no convergence argument is needed. Choosing the odd terms to be the naturals keeps it integral and non-trivial.
Generalization: for any sequence
b₁, b₂, b₃, ..., the sequenceb₁, -b₁², b₂, -b₂², b₃, -b₃², ...satisfies the property with sum 0. Rational family:b_k = 1/kgives1, -1, 1/2, -1/4, 1/3, -1/9, ....