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How fast a sequence can grow in $\ell_1$ norm before it diverges in $\ell_2$?

Let $\{a_n\} \in \ell_2(\mathbb{R})$, i.e. such that $\sum_{n=1}^\infty a_n^2 < \infty$, we want to find $\sum_{n=1}^N |a_n| = \Theta(N^{\varepsilon})$ (as $N \to \infty$) and $\varepsilon$ be the largest possible, what is the value of $\varepsilon$?

Some remarks:

  1. By Hölder inequality, we have a necessary condition that definetely, $$ cN^\varepsilon \le \sum_{k=1}^N |a_n| \le N^{1/2} \left(\sum_{k=1}^N a_n^2 \right)^{1/2} $$ that is $c^2 N^{-(1-2\varepsilon)} \le \sum_{n=1}^N a^2_n$, because the series in $\ell^2$ must converges, we have that $\varepsilon \le 1/2$.
  2. The sequence $a_n := (n+1)^{\varepsilon} - n^{\varepsilon}$ grows as $N^\varepsilon$ for $0 < \varepsilon < 1/4$, and it satisfies the conditions: $$ \sum_{n=1}^\infty |a_n| = +\infty$$ and $$\sum_{n=1}^\infty a^2_n \le \varepsilon^2 \sum_{n=1}^\infty ((n+1)^{2\varepsilon-2}) \le \zeta(2-2\varepsilon) < \infty$$ where we used Taylor approximation and the fact that the $\zeta(\beta)$ converges as $\beta = 2 - 2\varepsilon > 1$.

Finally, polynomial ''growth'' is the fastest growth possible or exponential (for very low parameters) is possible? (I conjecture that a growth of $\Theta(N^\varepsilon (\ln N)^\gamma)$ is possible by just adding some harmonic sequence to an already polynomial growing sequence, i.e. $\frac{(n+1)^{\varepsilon + \gamma} - n^{\varepsilon + \gamma}}{n^\eta}$ but I'm not so sure).

user807606
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  • You can get arbitrarily close to $\epsilon = 1/2$ by using $a_n=(1/n)^{1-\epsilon}$ for any $\epsilon \in (0, 1/2)$. – Michael Aug 02 '23 at 19:57
  • You can also use $a_n=\frac{1/\sqrt{n}}{\log(n+2)}$ to get $\sum_{n=1}^{\infty} a_n^2<\infty$ and $\sum_{n=1}^N a_n \geq \frac{\Theta(N^{1/2})}{\log(N+2)}$. This differs from your $\Theta(N^{1/2})$ upper bound by only a $\log(N+2)$ factor. – Michael Aug 02 '23 at 20:09
  • how do you prove that $a_n = 1/\sqrt{n}$ diverges as $N^{1/2}$ definetely? As far as I know the harmonic series, i.e. the series whose entries are $a_n = 1/n$, grows like $\ln(N)$. – user807606 Aug 02 '23 at 20:19
  • You compare to an integral: For $c>0$ we get $$\int_1^{N+1} (1/t)^cdt \leq \sum_{n=1}^N 1/n^c \leq 1 + \int_1^{N}(1/t)^cdt $$ [My original comment messed up the integration] – Michael Aug 02 '23 at 20:27
  • Right, seems fine to me. I initially excluded that kind of sequences because of the harmonic's series slow divergence. Then, one can also use $a_n := 1/\sqrt{n} \log(n+2)^{-1+\gamma}$ for $\gamma <1$ or add some $\log \log$ stuff I imagine, right? – user807606 Aug 02 '23 at 20:36
  • Yes, something like that. Related: https://math.stackexchange.com/questions/452053/is-there-a-slowest-rate-of-divergence-of-a-series – Michael Aug 02 '23 at 20:46

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