What A Series

$f(x) = (x+1)(x+2)(x+3)...(x+n)\\ \begin{aligned} \Rightarrow f'(x) & = [(x+2)(x+3)(x+4)...(x+n)] \\ & \quad \quad +[(x+1)(x+3)(x+4)...(x+n)]\\ & \quad \quad \quad + [(x+1)(x+2)(x+4)...(x+n)] +...\\ &\quad \quad \quad \quad ...+[(x+1)(x+2)(x+3)...(x+n-1)] \\ & = \dfrac {f(x)}{(x+1)} + \dfrac {f(x)}{(x+2)} + \dfrac {f(x)}{(x+3)} +...+ \dfrac {f(x)}{(x+n)} \\ &= f(x)\left(\frac{1}{x+1} + \frac{1}{x+2} + \frac{1}{x+3} +...+ \frac{1}{x+n}\right) \\ \Rightarrow f'(0) & = f(0)\left(\frac{1}{1} + \frac{1}{2} + \frac{1}{3} +...+ \frac{1}{n}\right) \\ & = \boxed{n! \left(1+ \dfrac{1}{2} + \dfrac{1}{3} +...+ \dfrac{1}{n}\right)} \end{aligned}$

Okay, let us try on the logarithm method.

$f(x)=\displaystyle \prod_{k=1}^n {(x+k)}$

$\begin{aligned} \Rightarrow \ln{(f(x))} & = \displaystyle \sum_{k=1}^n {\ln{(x+k)}} \\ \dfrac {f'(x)}{f(x)} & = \displaystyle \sum_{k=1}^n {\dfrac{1}{x+k}} \\ \Rightarrow f'(x) & = f(x) \displaystyle \sum_{k=1}^n {\dfrac{1}{x+k}} \\ \Rightarrow f'(0) & = f(0) \displaystyle \sum_{k=1}^n {\dfrac{1}{k}} = \boxed{n!\left( 1 + \dfrac{1}{2} + \dfrac{1}{3} +...+ \dfrac{1}{n} \right) } \end{aligned}$

$Ln(f(x)) = \displaystyle\sum_{k=1}^{n} (x+k)$ $\Rightarrow\frac{f'(x)}{f(x)} = \displaystyle\sum_{k=1}^{n} \frac{1}{x+k} \Rightarrow\ f'(x) = \displaystyle\sum_{k=1}^{n} \frac{f(x)}{x+k}$ $\therefore\ f'(0) = n! (1+\frac{1}{2} +\frac{1}{3} + \cdots + \frac{1}{n} )$

I have little familiarity with the logarithm method, but found another relatively simple solution.

Since we seek $f'(0)$ , we know that the only important terms of $f(x)$ are those that are part of the coefficient of $x$ , all others disappear or evaluate to $0$ .

So if we imagine that the entire sequence is multiplied out into individual terms, the important ones are $(x \cdot 2 \cdot 3 \cdot 4 \cdot ... \cdot n),$ $(1 \cdot x \cdot 3 \cdot 4 \cdot ... \cdot n),$ $(1 \cdot 2 \cdot x \cdot 4 \cdot ... \cdot n),$ ... all the way up to $(1 \cdot 2 \cdot ... \cdot (n-1) \cdot x).$

The $i$ th important term in the sequence equals $n!/i \cdot x$ .

When we take the derivative, the coefficients remain, and they add up to

$n!/1 + n!/2 + n!/3 + n!/4 + ... n!/n \\ = n!(1 + 1/2 + 1/3 + ... + 1/n).$

QED .

Note: Slightly more generally, to find $f'(k)$ (where $f$ is still the function above), we could probably use a transformation on $f$ , such as $f(x) = g(x - k)$ , allowing us to solve the problem in terms of $g'(0)$ . I haven't tried to see if this works (it's getting a little late in my time zone), so it is purely speculation.