The question arises from wiki link on the convergence of Fourier series.
Suppose $\{a_n\},n=1,2,3,\cdots$ is a real sequence and $\sum_{n}a_n^2$ converges.
Question: Cauchy-Schwarz implies that $$\sum_{n=1}^Na_n=O(N^{1/2}).$$ Is $O(N^{1/2})$ the best possible estimation for the partial sum of $\{a_n\}$? Can we improve it to $o(N^{1/2})$?
Yes, we can improve it to $o(N^{1/2}).$ Let $\epsilon>0.$ Then there exists $N_0$ such that
$$\left(\sum_{n=N_0+1}^{\infty}a_n^2\right)^{1/2}<\epsilon.$$
So for $N>N_0,$
$$\sum_{n=1}^{N}|a_n| = \sum_{n=1}^{N_0}|a_n| + \sum_{n=N_0+1}^{N}|a_n|$$ $$ \le \sum_{n=1}^{N_0}|a_n| + \left(\sum_{n=N_0+1}^{\infty}a_n^2\right)^{1/2}(N-N_0)^{1/2} \le \sum_{n=1}^{N_0}|a_n| +\epsilon\cdot N^{1/2}.$$
Therefore
$$\limsup_{N\to \infty}\frac{\sum_{n=1}^{N}|a_n|}{N^{1/2}} \le 0 + \epsilon = \epsilon.$$
Since $\epsilon$ is arbitrarily small, we see this $\limsup$ is $0.$ This proves $\sum_{n=1}^{N}|a_n|= o(N^{1/2})$ as desired.