Reciprocal Of Sum Of Reciprocals

Let $x = \dfrac{1}{\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}}$ .

Then we have: $\dfrac 1{\frac 7{2003}} < x < \dfrac 1{\frac 7{2009}} \implies 286.1428 < x < 287 \implies \lfloor x \rfloor = \boxed{286}$

I used differentials to approximate the value of the individual fractions in the denominator. Consider the Denominator

Let $f(x)=1/x$ . Then $f(x+\delta x) = f(x)+f^{'}(x) \cdot \delta x$ , where I've taken $\delta x$ to be a deviation about a central value $x$ .

So, $\frac{1}{x+\delta x} = \frac{1}{x} - \frac{1}{x^2} \delta x$ Now let us choose the central value $x$ to be $2006$ . Then the approximated denominator becomes:

$D = \sum_{i=1}^{7}{\frac{1}{2006} - \frac{1}{2006^2} \delta x_{i}}$ , since there are $7$ terms in the denominator. The second part of the summation is zero, as the sum of the deviations about $2006$ is zero. So, the approximated denominator is $D= \sum_{i=1}^{7} \frac{1}{2006} = \frac{7}{2006}$ .

Hence the answer is $\lfloor 1/D \rfloor = \lfloor \frac{2006}{7} \rfloor =\lfloor 286.5714 \rfloor = \boxed{286}$