 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Cube rooted till 2019 (Posted on 2019-10-23) 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 No Rating Comments: ( Back to comment list | You must be logged in to post comments.) 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) 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).

Then I + (1/2)*(1 + 1/cbrt) < 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 Please log in:

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