23. When $n=1,2$, direct verification shows that the inequality obviously holds. When $k \geqslant 3$,
$$\frac{1}{k^{3}}<\frac{1}{(k-1) k(k+1)}=\frac{1}{2}\left[\frac{1}{(k-1) k}-\frac{1}{k(k+1)}\right], k=3,4, \cdots, n$$

Adding up, we get
$$\sum_{k=3}^{n} \frac{1}{k^{3}}<\frac{1}{2}\left[\frac{1}{6}-\frac{1}{n(n+1)}\right]<\frac{1}{12}$$

Therefore,
$$\sum_{k=1}^{n} \frac{1}{k^{3}}<1+\frac{1}{8}+\frac{1}{12}=1+\frac{5}{24}<1+\frac{6}{24}=\frac{5}{4}$$