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Summing inverses II (Posted on 2012-03-16) Difficulty: 3 of 5
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.

No Solution Yet Submitted by K Sengupta    
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Solution solution | Comment 10 of 12 |
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 2012-03-17 13:47:03
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