# 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?

• 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 $0$. – Kavi Rama Murthy May 30 '20 at 0:04
• Is $a_i\ge 0$ in the hypothesis? – Gae. S. May 30 '20 at 0:10
• I suspect too but I can't prove it. Did you use integral test? – Yip Jung Hon May 30 '20 at 0:11
• Is the hypothesis that $\sum_{i=1}^\infty a_i$ converges, or that $\sum_{i=1}^\infty a_i^2$ converges? – Stefan Lafon May 30 '20 at 0:12
• @Gae.S.No, but it is implied since each $a_i$ is made out of a sum of squares – Yip Jung Hon May 30 '20 at 0:12

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.

• 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 – Yip Jung Hon May 30 '20 at 1:22
• Yup, it's simple squaring (and I was lazy to write it all out). The crux (of my proof) is the claim. – Calvin Lin May 30 '20 at 1:24
• noice this is pimpler – dezdichado May 30 '20 at 2:06

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}}

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.

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.