Limits question

**champrock** · April 4th 2009, 10:36 AM

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

**o_O** · April 4th 2009, 12:26 PM

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

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

You can see that since the degrees of the denominators is greater than that of the numerators, they dominate (for each term).

**awkward** · April 4th 2009, 01:43 PM

Originally Posted by o_O

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

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

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 $\sum_{i=1}^n \sqrt{i-1} = S_n$ and observe that

$\int_2^n \sqrt{x-2} \, dx< S_n < \int_1^n \sqrt{x-1} \, dx$
$(2/3) (n-2)^{3/2} < S_n < (2/3) (n-1)^{3/2}$
$\frac{(2/3) (n-2)^{3/2}}{n^{3/2}} < \frac{S_n}{n^{3/2} }< \frac{(2/3) (n-1)^{3/2}}{n^{3/2}}$

so, taking the limit,

$2/3 \leq \lim_{n \to \infty} \frac{S_n}{n^{3/2}} \leq 2/3$

**champrock** · April 4th 2009, 08:26 PM

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

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

**awkward** · April 5th 2009, 05:26 AM

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 $\sqrt{n-1}$ 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 $y = \sqrt{x-1}$ and $y = \sqrt{x-2}$ . $S_n$ is the total area of the first n rectangles. Can you "see" the inequality now?

**Danny** · April 5th 2009, 05:35 AM

Your limit is equal to a Riemann sum $\lim_{n \to \infty} \frac{\sum_{i=1}^{n} \sqrt{n-i}}{n^{\frac{3}{2}}} = \lim_{n \to \infty} \sum_{i=1}^{n} \sqrt{1-\frac{i}{n}} \cdot \frac{1}{n} = \int_0^1 \sqrt{1-x}\,dx$

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