\(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2013}-\dfrac{1}{2013}+\dfrac{1}{2015}\)
\(=1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{2015}-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2014}\right)\)
Thêm bớt \(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2014}\) vào tổng ta được:
\(S=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}+\dfrac{1}{2015}-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2014}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1007}\right)\)
\(=\dfrac{1}{1008}+\dfrac{1}{1009}+...+\dfrac{1}{2015}\)
\(\Rightarrow S=P\)
\(\Rightarrow S-P=0\)
\(\Rightarrow\left(S-P\right)^{2016}=0^{2016}=0\)
