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
$\displaystyle \lim_{n \to \infty} \frac{\sum_{i=1}^{n} \sqrt{n-1}}{n^{\frac{3}{2}}} = \lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + \cdots + (n-2)^{\frac{1}{2}} + (n-1)^{\frac{1}{2}}}{n^{\frac{3}{2}}} $
$\displaystyle = \lim_{n \to \infty} \left( \frac{\sqrt{1}}{n^{\frac{3}{2}}} + \frac{\sqrt{2}}{n^{\frac{3}{2}}} + \cdots + \frac{(n-2)^{\frac{1}{2}}}{n^{\frac{3}{2}}} + \frac{(n-1)^{\frac{1}{2}}}{n^{\frac{3}{2}}}\right)$
You can see that since the degrees of the denominators is greater than that of the numerators, they dominate (for each term).
Yes, but the number of terms in the series is n, which can be tricky...
Let $\displaystyle \sum_{i=1}^n \sqrt{i-1} = S_n$ and observe that
$\displaystyle \int_2^n \sqrt{x-2} \, dx< S_n < \int_1^n \sqrt{x-1} \, dx $
$\displaystyle (2/3) (n-2)^{3/2} < S_n < (2/3) (n-1)^{3/2}$
$\displaystyle \frac{(2/3) (n-2)^{3/2}}{n^{3/2}} < \frac{S_n}{n^{3/2} }< \frac{(2/3) (n-1)^{3/2}}{n^{3/2}}$
so, taking the limit,
$\displaystyle 2/3 \leq \lim_{n \to \infty} \frac{S_n}{n^{3/2}} \leq 2/3$
No, the sum is not equal to the integral.
Try this: Draw x and y coordinate axes. Sketch rectangles of height $\displaystyle \sqrt{n-1}$ 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 $\displaystyle y = \sqrt{x-1}$ and $\displaystyle y = \sqrt{x-2}$. $\displaystyle S_n$ is the total area of the first n rectangles. Can you "see" the inequality now?