# Does $\sum\limits_{k=1}^n a_k^2$ imply $\sum\limits_{l=1}^k a_k \in o(\sqrt{n})$?

I'm trying to determine some limits and it makes me wonder if my intuition about asymptotics is just wrong: Our calculus professor used to say that $\sum\limits_{n=1}^{\infty} \frac{1}{n}$ is basically the slowest divergent sum. Now, in my understanding this would mean that if I have $$\sum\limits_{n=1}^{\infty} a_n^2 < \infty$$ it would follow that $a_n^2 \in o(\frac{1}{n})$ and therefore $a_n \in o(\frac{1}{n^{1/2}})$. Since $\sum\limits_{k=1}^n \frac{1}{k^{1/2}} \sim \sqrt{n}$ it follows that $\sum\limits_{k =1}^n a_n \in o(\sqrt{n})$.

Am I using the Landau-symbols correctly and can one just carry out calculations with them like that?

You seem to have in mind sequences $a_n$ determined by smooth functions $f(x)$ evaluated at integers $n$, which I agree is a good mental picture; unfortunately, series don't have to be all that well behaved in general. Take for example $$a_n = \begin{cases} n^{-1/4}, &\text{if n is a power of 2}, \\ 0, &\text{otherwise.} \end{cases}$$ Then $\sum_{n=1}^\infty a_n^2$ converges (the nonzero terms form a geometric series), but $a_n^2$ is not $o(1/n)$.
A general moral might be: if the only information you have is about an aggregate statistic of a sequence (like an $O$-estimate for its sum), then you're not going to be able to deduce much information about the individual terms (like an $O$-estimate for each $a_n$) without additional information about the structure of the sequence.
I think there is a flaw in reasoning. There is no slowest divergent series. Suppose $a_n>0$ and $\sum a_n$ be divergent, one can prove that $\sum \frac{a_n}{1+a_n}$ is also divergent. Clearly every term of the latter series is smaller than the former, $\frac{a_n}{1+a_n}<a_n$.