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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?

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  • $\begingroup$ I suspect that $a_i$ must be $0$ for all $i$. I changed the summation to an integral - $f(x)=\int_x^{\infty} f(t)^{2} dt$ and found that $f$ must be $0$. $\endgroup$ – Kavi Rama Murthy May 30 '20 at 0:04
  • $\begingroup$ Is $a_i\ge 0$ in the hypothesis? $\endgroup$ – Gae. S. May 30 '20 at 0:10
  • $\begingroup$ I suspect too but I can't prove it. Did you use integral test? $\endgroup$ – Yip Jung Hon May 30 '20 at 0:11
  • $\begingroup$ Is the hypothesis that $\sum_{i=1}^\infty a_i$ converges, or that $\sum_{i=1}^\infty a_i^2$ converges? $\endgroup$ – Stefan Lafon May 30 '20 at 0:12
  • $\begingroup$ @Gae.S.No, but it is implied since each $a_i$ is made out of a sum of squares $\endgroup$ – Yip Jung Hon May 30 '20 at 0:12
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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.

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  • $\begingroup$ Right, and the last inequality is true because $\sqrt{1+\frac{4}{x}} > 1+\frac{2}{x+1}$, which can be seen by squaring both sides $\endgroup$ – Yip Jung Hon May 30 '20 at 1:22
  • $\begingroup$ Yup, it's simple squaring (and I was lazy to write it all out). The crux (of my proof) is the claim. $\endgroup$ – Calvin Lin May 30 '20 at 1:24
  • $\begingroup$ noice this is pimpler $\endgroup$ – dezdichado May 30 '20 at 2:06
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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.

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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.

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