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So of course it does not converge. The behavior is interesting however. Below is the graph of the $\log$ of the product going up to $5000.$ There can be runs very rich in square free integers with an even number of divisors which cause a dramatic shift. At $509$ the product is about $5\times 10^{-13}.$ Of the next $45$ square free integers ,$15$ provide a factor around $\frac{1}{500}$ and the other $30$ provide a factor around $500$ so the product at $586$ is around $1.7 \times 10^{23}.$ Of course there are bigger swings in both directions later on. This argues against any simple convergence acceleration used as smoothing.

Rearranging can do anything but a reasonable procedure might be to look at the $2^k$ square free integers divisible only by (some) of the first $k$ primes. The product over those comes out to be $1$ for $p_k \ge 3.$ So this could be taken as suggesting some balance. The partial product $\prod_{n=1}^N n^{\mu(n)}$ probably never has an even denominator for $5 \lt N \le 10000$ . However the numerator is just four times an odd number for $N=6590$ and $N=6593.$ So, if forced to guess, I'd guess that it is odd someplace(s) past $10000.$ Something similar seems to be happening with larger small primes.

So of course it does not converge. The behavior is interesting however. Below is the graph of the $\ln$ of the product going up to $5000.$ There can be runs very rich in square free integers with an even number of divisors which cause a dramatic shift. At $509$ the product is about $5\times 10^{-13}.$ Of the next $45$ square free integers ,$15$ provide a factor around $\frac{1}{500}$ and the other $30$ provide a factor around $500$ so the product at $586$ is around $1.7 \times 10^{23}.$ Of course there are bigger swings in both directions later on. This argues against any simple convergence acceleration used as smoothing.

Rearranging can do anything but a reasonable procedure might be to look at the $2^k$ square free integers divisible only by (some) of the first $k$ primes. The product over those comes out to be $1$ for $p_k \ge 3.$ So this could be taken as suggesting some balance. The partial product $\prod_{n=1}^N n^{\mu(n)}$ never has an even denominator for $5 \lt N \le 10000$ . However the numerator is just four times an odd number for $N=6590$ and $N=6593.$ So, if forced to guess, I'd guess that it is odd someplace(s) past $10000.$ Something similar seems to be happening with larger small primes.

So of course it does not converge. The behavior is interesting however. Below is the graph of the $\log$ of the product going up to $5000.$ There can be runs very rich in square free integers with an even number of divisors which cause a dramatic shift. AT $509$ the product is about $5\times 10^{-13}.$ Of the next $45$ square free integers ,$15$ provide a factor around $\frac{1}{500}$ and the other $30$ provide a factor around $500$ so the product at $586$ is around $1.7 \times 10^{23}.$ Of course there are bigger swings in both directions later on. This argues against any simply minded convergence acceleration used as smoothing.

Rearranging can do anything but a reasonable procedure might be to look at the $2^k$ square free integers divisible only by (some) of the first $k$ primes. The product over those comes out to be $1$ for $p_k \ge 3.$ So this could be taken as suggesting some balance. As far as $10000$ the product never has an even denominator after $5$. However the numerator is just four times an odd number for $6590,6593$ so if forced to guess I'd guess thatit is negative someplace(s) past $10000.$ Something similar seems to be happening with larger small primes.

So of course it does not converge. The behavior is interesting however. Below is the graph of the $\log$ of the product going up to $5000.$ There can be runs very rich in square free integers with an even number of divisors which cause a dramatic shift. At $509$ the product is about $5\times 10^{-13}.$ Of the next $45$ square free integers ,$15$ provide a factor around $\frac{1}{500}$ and the other $30$ provide a factor around $500$ so the product at $586$ is around $1.7 \times 10^{23}.$ Of course there are bigger swings in both directions later on. This argues against any simple convergence acceleration used as smoothing.

Rearranging can do anything but a reasonable procedure might be to look at the $2^k$ square free integers divisible only by (some) of the first $k$ primes. The product over those comes out to be $1$ for $p_k \ge 3.$ So this could be taken as suggesting some balance. The partial product $\prod_{n=1}^N n^{\mu(n)}$ probably never has an even denominator for $5 \lt N \le 10000$ . However the numerator is just four times an odd number for $N=6590$ and $N=6593.$ So, if forced to guess, I'd guess that it is odd someplace(s) past $10000.$ Something similar seems to be happening with larger small primes.

So of course it does not converge. The behavior is interesting however. Below is the graph of the $\log$ of the product going up to $5000.$ There can be runs very rich in square free integers with an even number of divisors which cause a dramatic shift. AT $509$ the product is about $5\times 10^{-13}.$ Of the next $45$ square free integers ,$15$ provide a factor around $\frac{1}{500}$ and the other $30$ provide a factor around $500$ so the product at $586$ is around $1.7 \times 10^{23}.$ Of course there are bigger swings in both directions later on. This argues against any simply minded convergence acceleration used as smoothing.

Rearranging can do anything but a reasonable procedure might be to look at the $2^k$ square free integers divisible only by (some) of the first $k$ primes. The product over those comes out to be $1$ for $p \ge 3$

So of course it does not converge. The behavior is interesting however. Below is the graph of the $\log$ of the product going up to $5000.$ There can be runs very rich in square free integers with an even number of divisors which cause a dramatic shift. AT $509$ the product is about $5\times 10^{-13}.$ Of the next $45$ square free integers ,$15$ provide a factor around $\frac{1}{500}$ and the other $30$ provide a factor around $500$ so the product at $586$ is around $1.7 \times 10^{23}.$ Of course there are bigger swings in both directions later on. This argues against any simply minded convergence acceleration used as smoothing.

Rearranging can do anything but a reasonable procedure might be to look at the $2^k$ square free integers divisible only by (some) of the first $k$ primes. The product over those comes out to be $1$ for $p_k \ge 3.$ So this could be taken as suggesting some balance. As far as $10000$ the product never has an even denominator after $5$. However the numerator is just four times an odd number for $6590,6593$ so if forced to guess I'd guess thatit is negative someplace(s) past $10000.$ Something similar seems to be happening with larger small primes.