# Find formula for sum of square roots

by zillac
Tags: formula, roots, square
 P: 4 1. The problem statement, all variables and given/known data 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) 2. Relevant equations Area under curve $$y = \sqrt{x}$$ in [0, 1] is $$\frac{1}{n^{\frac{3}{2}}}\sum^{n}_{i=1}\sqrt{i}$$ 3. 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.
 Sci Advisor HW Helper Thanks P: 25,228 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.
 P: 4 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.
 Sci Advisor HW Helper Thanks P: 25,228 Find formula for sum of square roots 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).
P: 270
 Quote by 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.
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.
HW Helper
Thanks
P: 26,148
 Quote by zillac 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.
P: 4
 Quote by quadraphonics 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 = $$\frac{\sqrt{n}}{n^\frac{3}{2}-(n-1)^\frac{3}{2}}$$
...
= $$\frac{1+(1-1/n)^\frac{3}{2}}{3-3/n-3/n^2}$$
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?
 Sci Advisor HW Helper Thanks P: 25,228 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.

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