By the Cauchy-Schwartz inequality, $$b_{n+1}^2 \ge \frac{b_1^2}{1^3}+\frac{b_2^2}{2^3}+...+\frac{b_n^2}{n^3} \ge \frac{(b_1+b_2+\cdots +b_n)^2}{1^3+2^3+\cdots+n^3}$$. Rearranging, we get that $\left(\frac{b_{n+1}}{b_1+b_2\cdots+ b_n}\right)^2 \ge \frac{1}{1^3+2^3+\cdots n^3}= \frac{1}{\left(\frac{n(n+1)}{2}\right)^2}= \left(\frac{2}{n(n+1)} \right)^2$. Hence, we get that $\frac{b_{n+1}}{b_1+b_2+\cdots+b_n} \ge \frac{2}{n(n+1)} \implies \sum_{n=1}^M\frac{b_{n+1}}{b_1+b_2+\cdots+b_n} \ge \sum_{n=1}^M \frac{2}{n(n+1)} = 2- \frac{2}{M+1}$. Now for sufficiently large $M$, it is clear that this expression will get arbitrarily close to $2$, and we are done.