jmc

By the change-of-base formula, the expression is equivalent to $$\frac{\log (a+1)}{\log a}\cdot \frac{\log (a+2)}{\log (a+1)}\cdots \frac{\log (b-1)}{\log (b-2)}\cdot \frac{\log b}{\log (b-1)}.$$Almost all the terms cancel, leaving just $$\frac{\log b}{\log a},$$which equals ${\log }_{a}b$ by the change-of-base formula again. Therefore, ${\log }_{a}b=2,$ so $b=a^2.$

We are given that the expression contains $870$ logarithms, so $(b-1)-a+1=870,$ or $b-a=870.$ Substituting $b=a^2$ gives $a^2-a=870,$ or $a^2-a-870=0,$ which factors as $(a-30)(a+29)=0.$ Since $a$ must be positive, we have $a=30,$ and so $b=a^2=900.$ Thus, $a+b=30+900=\boxed{930}.$