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Let me do the second one, it looks more interesting.
I hope you are familar with the Euler-Wallis product formula.
Thus,
If,
We have,
Thus,
Reciprocal time,
Factor,
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Now we do thy problem
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The limit of,
If you multiply the numerator and denominator by the missing terms in order to obtain the Euler-Wallis product you will have the expression above. But since is missing there will be an extra division so,
Now the Wallis product will converge to and thus, the product of these two convergents tells us the limit is zero.
Note, I was extremely lazy in checking whether its square will converge to Euler-Wallis product because the last term is missing. So I might be wrong and the limit is zero.
The idea is that we are going to assume that the Dirichlet function is Riemann integrable then arrive at a contradiction.
A bit gappy, but then so is my "proof", but I suppose because its mine
I know how to fill those gaps (and that they do fill).
(I agree that this is more interesting, took some thought to deduce
what the limit must be - always an advantage before starting a proof)
RonL
Do not think that I came up with this full solution be realizing that the solutions for the sine functions form the infinite product. I have seen similar ones before, in fact, when I saw the evens and odd seperated by division a thought immediately occured to me, Wallis infinite product! I just manipulated it so that it worked. I think it would have been cooler is the user asked to find,
In such a case the solution who have involved ! I was hoping for that badly, and assumed that was the case until I arrived that the conlusion that there is one more term in the demoninator which made a possibly amazing problem very boring.
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In fact one way Sirling's asympomatic approximation is derived is from what I said!