Bounds of sum by using integral Is it correct to say?

$$\int _1^{n+1}\:\frac{1}{x^3}dx\le \sum _{k=1}^n\:\frac{1}{k^3}\le 1+\:\int _1^n\:\frac{1}{x^3}dx$$

Therefore:

$$\sum _{k=1}^n\:\frac{1}{k^3}\le 1+\:\int _1^n\:\frac{1}{x^3}dx=\frac{3}{2}-\frac{1}{2n^2}\le \frac{3}{2}$$

And the same thing for the other side.
Yes, that's right. Since the system won't allow such a short answer, I'll add that indeed

$$\sum_{k=1}^\infty\frac1{k^3}=\zeta(3)\approx1.202\lt\frac32\;.$$
Yes, that's right. Since the system won't allow such a short answer, I'll add that indeed

$$\sum_{k=1}^\infty\frac1{k^3}=\zeta(3)\approx1.202\lt\frac32\;.$$