A finite summation involving $2013$

Can you help me compute the summation below? $$1+\frac{1}{2}+\cdots+\frac{1}{2013}+\frac{1}{1\cdot2}+\frac{1}{1\cdot3}+\cdots+\frac{1}{2012\cdot 2013}+\cdots+\frac{1}{1\cdot2\cdots2013}$$

Solution 1:

Hint:

Consider the polynomial $f(x)=\left(x+1\right)\left(x+\dfrac12\right)\ldots\left(x+\dfrac{1}{2013}\right)$.
Can you relate the value $f(1)$ with your sum?

Solution 2:

Here's how it works for $n=3$.

Find a common denominator:

$$\begin{align} 1 + S &= 1 + \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{1\cdot2} + \frac{1}{1\cdot3} + \frac{1}{2\cdot3} + \frac{1}{1\cdot2\cdot3}\\ &=\frac{1\cdot2\cdot3 + 2\cdot3 + 1\cdot3 + 1\cdot2 + 3 + 2 + 1 + 1}{1\cdot2\cdot3}\\ &= \frac{(1+1)(1+2)(1+3)}{1\cdot2\cdot3}\\ &= \frac{2\cdot3\cdot4}{1\cdot2\cdot3}\\ &= 4 \end{align}$$ So, $S = 3$.

Generally, $(1 + S)n! = (x + 1)(x + 2) \cdots (x + n)\big|_{x=1} = (n + 1)!$, which gives $S = n$.