Let S be the sum defined in the problem
Let I be the definite integral of 1/cbrt[x] on the interval (1,2019).
The quantity E1 = S - 1 is then a numeric calculation of integral I, by simply taking rectangles that are bounded below the curve. E1 will be a notable underestimate of I. Then E1 << I, which implies S << I + 1.
The quantity E2 = S - (1/2)*(1 + 1/cbrt[2019]) is then a numeric calculation of integral I, by using an extended trapezoidal rule. E2 will be an over estimate of I. Then E2 > I, which implies S > I + (1/2)*(1 + 1/cbrt[2019]).
Then I + (1/2)*(1 + 1/cbrt[2019]) < S << I + 1. Calculating the expressions into decimals yields 238.655 < S << 239.116.
Since the error is skewed away from the larger bound then I conclude floor[S] = 238.
Integrals
