I want to prove (or disprove) that
$$\sum_{i=1}^n\log_2 i = \Theta(n\log_2 n)\,,$$
but I totally get stuck with this example. May someone help me with that.
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Use the fact that $\log a + \log b = \log (ab)$. – adrianN Oct 25 '17 at 12:04
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You can use LaTeX directly here (just put in dollars as you usually would), so I replaced your image with text. – David Richerby Oct 25 '17 at 12:57
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Asymptotically,
$$\frac{1}{4}n\log_2n < \frac{1}{2}n(\log_2n-1) = \sum_{i=\frac{n}{2}}^{n}\log_2\frac{n}{2}<\sum_{i=1}^{n} \log_2i < \sum_{i=1}^{n} \log_2 n = n \log_2 n\,.$$
David Richerby
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Szymon Stankiewicz
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The LHS is $\log(n!)$, which is known, by Stirling's formula, to be asymptotic to $$\frac12\log(2\pi n)+n(\log(n) - 1).$$
You can conclude.