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 $a_{i-1} = 0, a_{i-2} = 0, \ldots$, 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?
 A: 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. 
A: So assume $a_1 > 0$. $\ln a_n = \ln a_{n+1} + \ln(a_{n+1}+1)<\ln a_{n+1} +a_{n+1}$ so $\dfrac{a_n}{a_{n+1}}<e^{a_{n+1}}.$
Therefore, 
$$a_{n+1} = a_1\prod_{i=1}^n\dfrac{a_{i+1}}{a_i}>a_1e^{-\sum_{i=1}^na_{i+1}}.$$ But this gives a lower bound: 
$$a_{n+1} > a_1 e^{-S}$$ 
if $a_1 > 0$ and their sum converges to a positive number $S>0,$ which is in return a contradiction.  
A: Yes, the zero sequence is the only one. Otherwise $a_n>0$ for all $n$ and $$a_n-a_{n+1}=a_{n+1}^2\implies\frac{1}{a_{n+1}}-\frac{1}{a_n}=\frac{1}{1+a_{n+1}}\underset{n\to\infty}{\longrightarrow}1,$$ now Stolz–Cesàro theorem implies $\lim\limits_{n\to\infty}na_n=1$. Thus, $\sum\limits_{n=1}^{\infty}a_n$ diverges.
