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Aug 17, 2020 at 15:00 comment added Iosif Pinelis @AlapanDas : I have added a response to your comment.
Aug 17, 2020 at 14:59 history edited Iosif Pinelis CC BY-SA 4.0
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Aug 17, 2020 at 2:43 comment added Alapan Das The asymptotic comes from the integral $$4^n\int_{0}^{1} \left(\prod_{k=1}^{n} \cos^2(\pi kt) \right)dt$$. But how to calculate the integral?
Aug 17, 2020 at 2:37 comment added Iosif Pinelis @LSpice : Thank you for your comment. I am sorry for the confusion. I was hesitating between $s_n$ (for the sum of the squares) and $a_n$ (the notation used on the linked page). Now this typo is fixed.
Aug 17, 2020 at 2:33 history edited Iosif Pinelis CC BY-SA 4.0
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Aug 17, 2020 at 1:06 comment added LSpice $a_n = s_n$, right?
Aug 17, 2020 at 0:48 history answered Iosif Pinelis CC BY-SA 4.0