We know that the prime number theorem is equivalent to the statement $$ M(x)=\sum_{n\le x}\mu(n)=o(x). $$ By using Ramanujan sums, we can write $M(x)$ as $$ M(x)=\sum_{q\le x}\sum_{\substack{0\lt a\le q \\ (a,q)=1}} e\Big(\frac{a}{q}\Big), $$ and this resembles some sums that appear in the proof of the large sieve. For example, we know that for any 1-periodic function $f$ with continuous derivative, we have $$ \sum_{q\le x}\sum_{\substack{0\lt a\lt q \\ (a,q)=1}}\Big|f\Big(\frac{a}{q}\Big)\Big|\le x^2\int_{0}^{1}|f(t)|dt + \int_{0}^{1} |f'(t)|dt. $$ Is there any way to adapt this type of machinery for $M(x)$?
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1$\begingroup$ Your second formula seems inconsistent with $M(1)=1$. Perhaps $0\lt a\lt q$ in your second formula should be $0\lt a\le q$ to account for the case $x=1$? $\endgroup$– Steven ClarkCommented Nov 14 at 15:45
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$\begingroup$ Yes that's right, I just edited the question, thank you! $\endgroup$– ItachiCommented Nov 14 at 20:08
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This isn't exactly along the lines you are suggesting, but Hildebrand does have a proof of the Prime Number Theorem which proceeds but estimating $M(x)$ using the large sieve inequality. See:
A. Hildebrand, The prime number theorem via the large sieve, Mathematika 33 (1986), 23-30.
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1$\begingroup$ I see, maybe at first it's not quite along the same lines but after giving it some thought I think this is kinda what I was looking for. I guess everytime we deal with exponential sums we are implicitly looking at divisibility conditions, so that the large sieve type inequality from that paper can be translated to an exponential sum type approach. Thank you $\endgroup$– ItachiCommented Nov 15 at 0:19