Verified answer

Let b(n) = SQRT( a(n) a(n+1) )

Then b(n+1) / b(n) = SQRT( a(n+1) a(n+2) ) / SQRT( a(n) a(n+1) ) = SQRT( a(n+2) / a(n) )

This ratio is less than 1 because a(n) has to be a decreasing sequence. Therefore b(n) converges.