Question:

Step 1: Find the general term. The numerator of the $r$-th term is $1+2+\cdots+r = \frac{r(r+1)}{2}$. The denominator is $1^3+2^3+\cdots+r^3 = \left(\frac{r(r+1)}{2}\right)^2$. Step 2: The $r$-th term is $T_r = \dfrac{\frac{r(r+1)}{2}}{\left(\frac{r(r+1)}{2}\right)^2} = \dfrac{2}{r(r+1)} = 2\left(\dfrac{1}{r} - \dfrac{1}{r+1}\right)$. Step 3: Sum: $S_n = 2\sum_{r=1}^{n}\left(\frac{1}{r} - \frac{1}{r+1}\right) = 2\left(1 - \frac{1}{n+1}\right) = \frac{2n}{n+1}$. Step 4: Since $S_n = \frac{2n}{n+1} = 2 - \frac{2}{n+1} < 2$ for all $n \geq 1$. Therefore $S_n < 2 < 3 < 4$. Step 5: $S_n$ is always less than 2, 3, and 4. It is NOT always less than 1 (e.g., $S_1 = 1$, $S_2 = 4/3 > 1$). So options (b), (c), (d) are correct.