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I need to prove that: $\sum_{k=0}^n k^{0.5} = \Theta (n^\frac {3}{2})$. I tried to evaluate it with integral because $ k^{0.5}$ is monotonically increasing. $\int_{0}^n k^{0.5} \le \sum_{k=1}^n k^{0.5} \le \int_{1}^{n+1} k^{0.5}$.

the left side is easy and give the solution of $ \frac {n^{1.5}}{1.5}$ which says that its low bound with $c=1/1.5$ I'm having troubles with the right side, I can't find the right constant for upper bound.

RobPratt
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shazar123
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  • You can use the answers to this post: https://math.stackexchange.com/a/422720/356329 – OmG Aug 09 '20 at 16:27
  • I saw this but I still didnt learn about Euler-Maclaurin. I need to use integrals. – shazar123 Aug 09 '20 at 16:29
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    Integrate like you did for the left, then factor an $n$ out of the term (you end up with $\frac23\cdot n^{3/2}\left(1+\frac1n\right)^{3/2}$. This makes it clear that any number greater than $\frac23$ suffices simply by taking $n$ large enough. – Clayton Aug 09 '20 at 16:43

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Actually your sum $S_n=\sum_{k=1}^n\sqrt{k}$ is equivalent to $n^{3/2}/(3/2)$ (by your argument), hence if $u_n\sim v_n$ then $\frac{u_n}{v_n}\rightarrow 1$ then the quotient is bounded since it is convergent so there is a lower and an upper bound for this quotient giving you the result that $S_n=\Theta(n^{3/2})$.

BlueCharlie
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