We can easily prove that P>0.

We have: \(P=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2013^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2012.2013}\)

                                                                 \(=1-\dfrac{1}{2013}< 1\)

\(\Rightarrow0< P< 1\)

So P isn't a integer.

We can easily prove that P>0.

We have: P=122+132+...+120132<11.2+12.3+...+12012.2013

                                                                 =1−12013<1

⇒0<P<1

So P isn't a integer.