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

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

My intuition is that the only sequence of this form is the zero sequence.

Here's what I have so far: $a_n - a_{n+1} = a_{n+1}^2 \implies a_{n+1} = \sqrt{a_n + \frac{1}{4}} - \frac{1}{2}$, but it doesn't seem to lead me anywhere.

Another line of thought is that if $a_i = 0$ for some $i$, it means that $\sum_{k=i+1}^\infty a_k^2=0$, which means that $a_k = 0$ for $k > i$. This will also mean $a_{i-1} = 0, a_{i-2} = 0...$, making the whole sequence the zero sequence.

It means that $a_i >0 $ for all $i$, yet $\lim a_i = 0$.

The last line I've tried is $a_1 = a_2^2 + a_3^2 + a_4^2 + ..., a_2 = a_3^2 + a_4^2 + ....$, so $\sum_{i=1}^\infty a_i = a_2^2 + a_3^2 + a_4^2 + ... +a_3^2 + a_4^2 + .... = a_2^2 + 2a_3^2 + 3a_4^2 = \sum_{i=2}^\infty (i-1)a_i^2$, which implies a stronger condition of having $ia_i^2 \to 0$. I'm hoping to get a contradiction but it doesn't seem to work.

Python seems to suggest that $(a_n) \approx \frac{1}{n}$ for large $n$.

Any hints?

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