If the value of

\(\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2014}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{2015}\right)\) \(-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2015}\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{2014}\right)\)

is \(\frac{a}{b}\) for relatively prime \(a\) and \(b\), then the value of \(a+b\) is

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