Find all sequences that has $\sum_{i=1}^\infty a_i$ converges, where $a_i = \sum_{k=i+1}^\infty a_k^2$.

Claim: If $a_n > \frac{1}{k}$, then $ a_{n+1} > \frac{1}{k+1}$.

Proof: Verify that for $ k > 0$,

$$ a_{n+1} = \frac{ - 1 + \sqrt{ 1 + 4 a_n } }{2} > \frac{ - 1 + \sqrt{ 1 +\frac{4}{k} } }{2} > \frac{ 1}{k+1}. $$

Corollary: If $ a_1 > \frac{1}{k} $, then $ \sum a_n > \sum \frac{1}{k - 1 + n }$ which diverges.
Hence, the only sequence where $ \sum a_n$ converges is the all-0 sequence.