\[\sum_{n=1}^\infty{\frac{1}{a_{n}^{3}}}=\sum_{k=1}^{\infty }\sum_{n=k^{2}}^{k^{2}+2k}\frac{1}{a_{n}^{3}}=\sum_{k=1}^{\infty}\frac{2k+1}{k^{3}}=\frac{\pi^{2}}{3}+\zeta (3).\]

You used the next smaller integer, not the closest one.

Just needs some fixing up: $\sum_{n=1}^\infty{\frac{1}{a_{n}^{3}}}=\sum_{k=1}^{\infty }\sum_{n=k^{2}-k+1}^{k^{2}+k}\frac{1}{a_{n}^{3}}=\sum_{k=1}^{\infty}\frac{2k}{k^{3}}=\frac{\pi^{2}}{3}$