\(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{2013}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2013}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{1006}\right)\)
\(=\dfrac{1}{1007}+\dfrac{1}{1008}+....+\dfrac{1}{2013}=P\)
Vậy \(S=P\)