# A finite summation involving $2013$

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

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I'm really not clear about what is the general term here. – Asaf Karagila Mar 18 '13 at 21:24
Just to clarify (because the $\frac{1}{1\cdot 3}$ term is perhaps wrong), we're adding up the sums of the inverse of the product of $n$ sequential elements for all $1\leq n \leq 2013$? – Ian Coley Mar 18 '13 at 21:25
So far (i.e., before corrections), this looks like a nightmare-exercise... – DonAntonio Mar 18 '13 at 21:26
I have seen this question somewhere on M.SE – Santosh Linkha Mar 18 '13 at 21:31
If having $\{1,2,3\}$ then $S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{1\cdot2}+\frac{1}{1\cdot3}+\frac{1}{2\cdot‌​3}+\frac{1}{1\cdot2\cdot3}$ – Saul of Tarsus Mar 18 '13 at 21:34

Hint:

Consider the polynomial $f(x)=\left(x+1\right)\left(x+\dfrac12\right)\ldots\left(x+\dfrac{1}{2013}\right)$.
The sum you are looking for is $f(1)-1$.

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+1 really nice one – Dominic Michaelis Mar 18 '13 at 21:49

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 $S = n$.

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Let denote by $$P=(x-2013)(x-2012)\cdots(x-1)=x^{2013}+a_{2012}x^{2012}+\cdots+a_0$$ then we know the relation between the coefficients $a_i$ and the roots: $$\sigma_k=(-1)^ka_{2013-k}.$$ Now it's easy to see that the given sum is equal to $$\frac{\sigma_{2012}+\cdots+\sigma_1+1}{\sigma_{2013}}.$$

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