Determine ⌈N⌉, when it is known that:
1 1 1
N = ------- + ------- + ... + ----------
⌈(1)2/3⌉ ⌈(2)2/3⌉ ⌈(2023)2/3⌉
Note: ⌈p⌉ is the ceiling of p, which is equal to the least integer greater than or equal to p.
I dropped the first term in the sum (which is just 1), so I didn't have to worry about improper integrals. The I constructed this compound inequality:
Integ {1 to 2023} dx/x^(2/3)
> Integ {1 to 2023} dx/ceil(x^(2/3))
> Sum {2 to 2023} 1/ceil(x^(2/3))
> Integ {2 to 2024} dx/ceil(x^(2/3)+1)
> Integ {2 to 2024} dx/(x^(2/3)+1)
The first and fourth parts of inequality come from converting the ceil function into a continuous function I know is greater/less than the respective ceil function. The second and third parts of the inequality come from taking left-hand and right-hand numeric integration, which are under and over estimates.
Then evaluating the leftmost and rightmost integrals.
1/x^(2/3) has an easy antiderivative of 3*cbrt(x), so that integral evaluates to 34.942.
1/(x^(2/3)+1) is trickier to evaluate, but starting with a u-substitution u=cbrt(x) and then partial fraction decomposition and other manipulation eventually gets the antiderivative of 3*(cbrt(x)-arctan(cbrt(x))); thus this integral evaluates to 32.392.
So then I come down to 34.942 > N-1 > 32.392. Adding the first term of 1 back in, then 35.942 > N > 33.392. So at this point I can definitely say ceil(N) is either 34, 35, or 36.
I'm gonna need something a bit more refined to narrow it down further to the answer of 35 that Charlie calculated by brute force.
Analytic attempt
