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Hello i have to solve the following problem find an $\displaystyle f(k)$ where $\displaystyle S_k=\theta(f(k))$ where $\displaystyle S_k =\sum_{n=1}^{k^2-1} \sqrt{n}$

I tried first of all to calculate or "limit" my sum using integral rule so i came up with $\displaystyle \frac{2(k^2-1)^{3/2}}{3} \leq S_k \leq \frac{2(k^3-1)}{3}$

but after that i am in a dead end as i do not know $\displaystyle S_k$ so i can not simplify anything can anyone help how to proceed this problem ? thanks

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There is a question on this already: – Aryabhata Nov 10 '10 at 21:05

As far as I see you already have the solution there. In Theta-notation only the fastest growing terms are important. And that is the same for the upper and lower bound.

In other words, you have:

$$\Theta(k^3) \leq S_k \leq \Theta(k^3).$$

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i do not think that i can say that as the (2/3)*(k^2-1)^(3/2) is not in a form k^3*c – ECE Nov 10 '10 at 21:17
Yes you can. $2/3(k^2 - 1)^{3/2}$ is the same as $(k^2)^{3/2} = k^3$ plus some low order terms, but we don't care about them in Theta notation. – user915 Nov 10 '10 at 21:23
Can someone agree with me here? So big_bang sees I'm right and he/she can ponder on it more. :D Also, look up the definition of Theta big_bang! – user915 Nov 10 '10 at 21:25
@big_bang: Michael is right. $2(k^2-1)^{3/2}/3 = \theta(k^3)$ – Aryabhata Nov 10 '10 at 21:27
Thanks Moron. You went wild there on the page with the similar question. :) – user915 Nov 10 '10 at 21:30

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