Does this series diverge if $a_n\to 0$?

Expanding on Winther's comment, for $x\ge3$, $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\frac{x}{\log(x)} &=\frac{\log(x)-1}{\log(x)^2}\\ &=\frac1{\log(x)}\left(1-\frac1{\log(x)}\right)\\ &\ge\frac1{\log(x)}\left(\frac{\log(3)-1}{\log(3)}\right)\\ \end{align} $$ Thus, $$ \begin{align} \sum_{k=4}^n\frac1{\log(k)} &\le\int_3^n\frac{\mathrm{d}x}{\log(x)}\\ &\le\int_3^n\frac{\mathrm{d}x}{\log(x)}\\ &\le\frac{\log(3)}{\log(3)-1}\left[\frac{x}{\log(x)}\right]_3^n\\ &\le\frac{\log(3)}{\log(3)-1}\frac{n}{\log(n)} \end{align} $$ Therefore, $$ \sum_{n=4}^N\frac{\frac1{\log(n)^2}}{\sum\limits_{k=4}^n\frac1{\log(k)}} \ge\frac{\log(3)-1}{\log(3)}\sum_{n=4}^N\frac1{n\log(n)} $$ which diverges.

Consider $a_n = \dfrac1{\log^s(n+1)}$ for $s>0$. We then have the $n^{th}$ terms of your new sequence to be $$b_n = \dfrac{a_n^2}{a_1+a_2+\cdots+a_n} = \dfrac1{\log^{2s}(n+1) \left(\displaystyle\sum_{k=1}^n \dfrac1{\log^s(k+1)}\right)}$$ Further, we have $$\displaystyle \sum_{k=1}^n \dfrac1{\log^s(k+1)} \leq \dfrac{n}{\log^s(2)}$$ Hence, we have $$b_n \geq \dfrac{\log^s(2)}{n\log^{2s}(n+1)}$$ Now if $s \leq 1/2$, we see that $\displaystyle \sum_{n=1}^{\infty} \dfrac{\log^s(2)}{n\log^{2s}(n+1)}$ diverges. Hence, choosing $a_n= \dfrac1{\log^s(n+1)}$ for $s \in (0,1/2]$ ensure that $a_n \to 0$, whereas $\displaystyle \sum \dfrac{a_n^2}{a_1+a_2+\cdots+a_n}$ diverges.