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I seek the asymptotic behavior for the following sum: $$\sum_{i=1}^{\lceil \frac{\log n}{1/2\log\log n} \rceil}\sqrt{\log(n/(\log n)^{i/2})}$$

Is there a better (and still simple) estimate than $O\left(\frac{\log^{3/2}n}{\log \log n}\right)$?

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    Are you sure that you want the ceiling, rather than the floor in that upper limit? Once $i$ gets above that expression, the log in the sqrt will be negative. – 2017-02-10
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    Yes, that was the floor. Thank you. – 2017-02-17

1 Answers 1

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Playing around to see what happens.

$S(n) =\sum_{i=1}^{\lfloor \frac{\log n}{1/2\log\log n} \rfloor}\sqrt{\log(n/(\log n)^{i/2})} $.

Note that floor is needed in the upper limit, otherwise the log can go negative.

$f(n) =\log(n/(\log n)^{i/2}) =\log(n)-(i/2)\log\log(n) $.

Let $a = \frac{\log n}{1/2\log\log n} $ and $b = \lfloor a \rfloor $.

$\begin{array}\\ \sqrt{f(n)} &= \sqrt{\log(n)-(i/2)\log\log(n)}\\ &= \sqrt{\log(n)}\sqrt{1-(i/2)\log\log(n)/\log(n)}\\ &= \sqrt{\log(n)}\sqrt{1-i/a}\\ \end{array} $

so that

$\begin{array}\\ S(n) &=\sqrt{\log(n)}\sum_{i=1}^{\lfloor a\rfloor}\sqrt{1-i/a}\\ &=\sqrt{\log(n)}a\sum_{i=1}^{\lfloor a\rfloor}\frac1{a}\sqrt{1-i/a}\\ &\approx\sqrt{\log(n)}a\int_0^1 \sqrt{1-x}dx\\ &=\sqrt{\log(n)}\dfrac{\log n}{1/2\log\log n}\int_0^1 \sqrt{x}dx\\ &=\dfrac{\log^{3/2} n}{1/2\log\log n}\dfrac{x^{3/2}}{3/2}\big|_0^1\\ &=\dfrac{4\log^{3/2} n}{3\log\log n}\\ \end{array} $

It looks like your big-oh estimate is correct, so nothing better can be done.