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 $a_{i-1} = 0, a_{i-2} = 0, \ldots$, 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 $a_1 = a_2^2 + a_3^2 + a_4^2 + \ldots, a_2 = a_3^2 + a_4^2 + \ldots$, so $\sum_{i=1}^\infty a_i = a_2^2 + a_3^2 + a_4^2 + \ldots + a_3^2 + a_4^2 + \ldots = a_2^2 + 2a_3^2 + 3a_4^2 = \sum_{i=2}^\infty (i-1)a_i^2$, 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?
Best Answer
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.