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Cube rooted till 2019 (Posted on 2019-10-23) Difficulty: 3 of 5
Find the floor of the given expression:

1/(1)1/3+1/(2)1/3+1/(3)1/3+...+1/(2019) 1/3

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Integrals Comment 3 of 3 |
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.

  Posted by Brian Smith on 2019-10-26 08:58:46
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