Find formula for sum of square roots

[zillac]

Homework Statement

Find the asymptotic formula for

\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{3}+...+\sqrt{n}

in the form of c \ast n^{\alpha}
Identify c and alpha.
(Do NOT use the fundamental theorem of calculus)

Homework Equations

Area under curve y = \sqrt{x} in [0, 1]
is \frac{1}{n^{\frac{3}{2}}}\sum^{n}_{i=1}\sqrt{i}

The Attempt at a Solution

...I'm really not sure from where to start...
I know c = \frac{2}{3} and \alpha = \frac{3}{2}, but do not know how to prove it.

 

[Dick]

Go with the area under the curve idea. The sum from 1 to n of sqrt(n) is approximated by the area under the curve sqrt(x) from x=0 to n. Set that up as an integral and evaluate it.

 

[zillac]

Sorry I forget to say that the fundamental theorem of calculus cannot be used, otherwise it is pretty straight forward.
Thanks for the quick reply.

 

[Dick]

Ok, then. If c is the area under the curve sqrt(x) on [0,1] and I(n) is your sum of square roots, then you have c=limit n->infinity (1/n^(3/2))*I(n) from your Riemann integral type expression (were you just given that?). So for large n, I(n)~c*n^(3/2). There's the 3/2. The easy way to get c is to integrate to get the area. But not allowed? Ok, then look at I(n)-I(n-1)~c*(n^(3/2)-(n-1)^(3/2)). Separate the n's from the c and take the limit as n -> infinity. Is that allowed? (All of this is really just doing calculus the HARD WAY).

 

[quadraphonics]

Sorry I forget to say that the fundamental theorem of calculus cannot be used, otherwise it is pretty straight forward.
Thanks for the quick reply.

Well, all the fundamental theorem of calculus says is that definite integrals can be written in terms of antiderivatives, so as long as you don't use that fact, it should be a problem, no? So, as Dick suggests, set up the problem as an integral, and then evaluate the Riemann sums directly and take the limit, rather than plugging in known antiderivatives.

 

[tiny-tim]

Welcome to PF!

Find the asymptotic formula for

\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{3}+...+\sqrt{n}

in the form of c \ast n^{\alpha}
Identify c and alpha.
(Do NOT use the fundamental theorem of calculus)

Hi zillac! Welcome to PF!

Hint: when n -> ∞, cn^α + √n -> c(n+1)^α.

 

[zillac]

Well, all the fundamental theorem of calculus says is that definite integrals can be written in terms of antiderivatives, so as long as you don't use that fact, it should be a problem, no? So, as Dick suggests, set up the problem as an integral, and then evaluate the Riemann sums directly and take the limit, rather than plugging in known antiderivatives.

I tried using Riemann sums and it will go back to finding the sum of the sequence (\frac{1}{n^{\frac{3}{2}}}\sum^{n}_{i=1}\sqrt{i})

I(n)-I(n-1)~c*(n^(3/2)-(n-1)^(3/2)) will work when alpha is found.
So,
c = \frac{\sqrt{n}}{n^\frac{3}{2}-(n-1)^\frac{3}{2}}
...
= \frac{1+(1-1/n)^\frac{3}{2}}{3-3/n-3/n^2}<br />
lim c = 2/3
---------
then for the first half of question
lim Area = lim 1/(n^3/2) * ∑√i ( lim n -> infinity)
-> Area * n^3/2 = ∑√i = c * n^a
-> alpha = 3/2..

Is this precise enough to conclude the value of alpha?

 

[Dick]

Sure, if you presume that the area is a well defined constant, the riemann sum immediately gives you the n^(3/2) dependence. You just have to work a little harder to get the c.