Question

Sol. [Key Concept: Evaluate the sum.]

$$f(x) = \sum_{r = 1}^{n}\mspace{2mu}\frac{rx^{r - 1}}{(x + 1)(x + 2)\ldots(x + r)}$$

$$\begin{matrix} f(x) = & \left( 1 - \frac{x}{1 + x} \right) + \sum_{r = 2}^{n}\mspace{2mu}\mspace{2mu}\frac{(x + r)x^{r - 1} - x^{r}}{(x + 1)(x + 2)\ldots(x + r)} \\ = & \left( 1 - \frac{x}{1 + x} \right) + \sum_{r = 2}^{n}\mspace{2mu}\mspace{2mu}\left( \frac{x^{r - 1}}{(x + 1)(x + 2)\ldots(x + r - 1)} - \frac{x^{r}}{(x + 1)(x + 2)\ldots(x + r)} \right) \\ = & 1 - \frac{x^{n}}{(x + 1)(x + 2)\ldots(x + n)} \\ 1 - f(x) = & \ \prod_{1 - \frac{x}{x + r}}^{n}\mspace{2mu}\mspace{2mu} \\ \Rightarrow & \frac{- f^{'}(x)}{1 - f(x)} = \frac{1}{x}\sum_{r = 1}^{n}\mspace{2mu}\mspace{2mu}\frac{r}{x + r} \Rightarrow f^{'}(x) = - \frac{x^{n - 1}}{(x + 1)(x + 2)\ldots(x + n)}\sum_{r = 1}^{n}\mspace{2mu}\mspace{2mu}\frac{r}{x + r} \end{matrix}$$

✘ (a) not the closed form

✔ (b) $f(x) = 1 - \frac{x^{n}}{(x + 1)\cdots(x + n)}$

✘ (c) wrong derivative

✔ (d) $f'(x) = - \frac{x^{n - 1}}{(x + 1)\cdots(x + n)}\sum_{r = 1}^{n}\frac{r}{x + r}$