# Limits question

• Apr 4th 2009, 10:36 AM
champrock
Limits 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
• Apr 4th 2009, 12:26 PM
o_O
$\displaystyle \lim_{n \to \infty} \frac{\sum_{i=1}^{n} \sqrt{n-1}}{n^{\frac{3}{2}}} = \lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + \cdots + (n-2)^{\frac{1}{2}} + (n-1)^{\frac{1}{2}}}{n^{\frac{3}{2}}}$

$\displaystyle = \lim_{n \to \infty} \left( \frac{\sqrt{1}}{n^{\frac{3}{2}}} + \frac{\sqrt{2}}{n^{\frac{3}{2}}} + \cdots + \frac{(n-2)^{\frac{1}{2}}}{n^{\frac{3}{2}}} + \frac{(n-1)^{\frac{1}{2}}}{n^{\frac{3}{2}}}\right)$

You can see that since the degrees of the denominators is greater than that of the numerators, they dominate (for each term).
• Apr 4th 2009, 01:43 PM
awkward
Quote:

Originally Posted by o_O
$\displaystyle \lim_{n \to \infty} \frac{\sum_{i=1}^{n} \sqrt{n-1}}{n^{\frac{3}{2}}} = \lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + \cdots + (n-2)^{\frac{1}{2}} + (n-1)^{\frac{1}{2}}}{n^{\frac{3}{2}}}$

$\displaystyle = \lim_{n \to \infty} \left( \frac{\sqrt{1}}{n^{\frac{3}{2}}} + \frac{\sqrt{2}}{n^{\frac{3}{2}}} + \cdots + \frac{(n-2)^{\frac{1}{2}}}{n^{\frac{3}{2}}} + \frac{(n-1)^{\frac{1}{2}}}{n^{\frac{3}{2}}}\right)$

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,

$\displaystyle 2/3 \leq \lim_{n \to \infty} \frac{S_n}{n^{3/2}} \leq 2/3$
• Apr 4th 2009, 08:26 PM
champrock
why did you take Sn less than "Integrate of (n-1)" ?

Shouldnt it be equal to Integrate of (n-1) itself?
• Apr 5th 2009, 05:26 AM
awkward
Quote:

Originally Posted by champrock
why did you take Sn less than "Integrate of (n-1)" ?

Shouldnt it be equal to Integrate of (n-1) itself?

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?
• Apr 5th 2009, 05:35 AM
Jester
Your limit is equal to a Riemann sum $\displaystyle \lim_{n \to \infty} \frac{\sum_{i=1}^{n} \sqrt{n-i}}{n^{\frac{3}{2}}} = \lim_{n \to \infty} \sum_{i=1}^{n} \sqrt{1-\frac{i}{n}} \cdot \frac{1}{n} = \int_0^1 \sqrt{1-x}\,dx$