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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: http://math.stackexchange.com/questions/5676/2010-11-10

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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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    Thanks Moron. You went wild there on the page with the similar question. :)2019-01-27