# Anyone going to help me expand?

Algebra Level 3

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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