Find the sum...

Consider f(x) = 1/√x
f'(x) = -1/2x√x is a decreasing function for x > 0
Hence the integral A = ∫ f(x) dx from 1 to 361 can be estimated by the rectangular estimation.
1/√2+1/√3+...+1/√361 < A < 1/√1+1/√2+1/√3+...+1/√360
Direct integration yields ∫ f(x) dx = 2√x from 1 to 361 = 38 - 2 = 36
So 1/√2+1/√3+1/√4+...+1/√361 < 36 ... (1)
and 36 < 1/√1+1/√2+1/√3+...+1/√360 ... (2)
(1) => 1/√1 +1/√2 +1/√3+1/√4+...+1/√361 < 37
(2) => 36 + 1/19 < 1/√1+1/√2+1/√3+...+1/√360 + 1/√361
So the integral part of 1/√1 +1/√2 +1/√3+1/√4+...+1/√361 is 36 and the square root is 6