The sum of the reciprocal of the square root of all the positive integers up to n is denoted by F(n), that is:
F(n) = 1+1/√2 + 1/√3 +...+ 1/√n
Determine the maximum value of n such that the integer part of the base ten expansion of F(n) DOES NOT exceed 2012.
*** For an extra challenge, solve this puzzle without using a computer program.
From what I know, finding the positive integer value of n without a computer program is difficult or very tedious. Finding an approximate value is less so. This can be done with using antiderivatives.
Integral from a to b of f(x)dx = F(b)  F(a)
f(x) = x^(1/2)
F(x) = x^(1/2 + 1)/(1/2 + 1) = 2*[x^(1/2)] = 2*SQRT(x)
F(a) = F(1) = 2*SQRT(1) = 2
F(b) = F(n) = 2*SQRT(n)
F(b)  F(a) = 2*SQRT(n)  2
2012 <= 2*SQRT(n)  2
2014 <= 2*SQRT(n)
1007 <= SQRT(n)
1014049 <= n
2013 > 2*SQRT(n)  2
2015 > 2*SQRT(n)
1007.5 > SQRT(n)
1015056.25 > n
1014049 <= n <= 1015056
Using a computer program finds the integer value to be
1014512. This value does indeed fall within the given range, yet it means that the approximate value 1014049 is about 0.046% short of the correct value. I am interested in knowing what more accurate method is proposed for the extra challenge.

Posted by Dej Mar
on 20120317 13:47:03 