Limit[(Sum[i^0.5, {i, n}] - n^0.5)/n^1.5, n -> \[Infinity]]

F[x] = [ 1^0.5 + 2^0.5 ....... (n-1)^0.5 ] / n*n^0.5

Find Limit f(x) where n tends to infinity.

Thanks

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- April 4th 2009, 11:36 AMchamprockLimits question
Limit[(Sum[i^0.5, {i, n}] - n^0.5)/n^1.5, n -> \[Infinity]]

F[x] = [ 1^0.5 + 2^0.5 ....... (n-1)^0.5 ] / n*n^0.5

Find Limit f(x) where n tends to infinity.

Thanks - April 4th 2009, 01:26 PMo_O

You can see that since the degrees of the denominators is greater than that of the numerators, they dominate (for each term). - April 4th 2009, 02:43 PMawkward
- April 4th 2009, 09:26 PMchamprock
why did you take Sn less than "Integrate of (n-1)" ?

Shouldnt it be equal to Integrate of (n-1) itself? - April 5th 2009, 06:26 AMawkward
No, the sum is not equal to the integral.

Try this: Draw x and y coordinate axes. Sketch rectangles of height for n = 1, 2, 3, 4, 5 (say), where the base of each rectangle is the line segment from (n, 0) to (n+1, 0). Now sketch the curves and . is the total area of the first n rectangles. Can you "see" the inequality now? - April 5th 2009, 06:35 AMJester
Your limit is equal to a Riemann sum