How to determine such slopes that are sufficient to assume that a sequence converges, and possibly use such slopes to determine the limit?
Here's n/((n#)^(1/n))
Can someone please guide me how to determine the limit?
By the way, a similar sequence with the factorial in place of the primorial will yield e.
Can the slope be used to determine whether a sequence very likely converges to zero or a positive value?
Does it look like n/((n#)^(1/n)) > 0 for all n?
What does theory on growth of primes say about whether or not (n#)^(1/n) grows quickly enough to make n/((n#)^(1/n)) null?
>>7693236
ln(n#)~n
n#~e^n
n#^/(1/n)~e
n/e∞
ln(p_n#)~nln(n)
p_n#~ne^n
p_n#^/(1/n)~e(n)^1/n
n^(1-1/n)/e∞
Isn't this really interesting, /sci/?
>>7693329
What?
>>7693325
Shouldn't this be fairly simplistic for you to answer, /sci/?
>>7693400
[math]\ln{(x\#)}= \sum_{p\leq{x}}^{ }{\ln{p}}[/math]
Prime number theorem states [math]\sum_{p\leq{x}}^{ }{\ln{p}}\sim{x}[/math] and [math]p_n\sim{n\ln{n}}[/math]
so, substituting into that first equation, [math]\ln{(p_n\#)}\sim{p_n}\sim{n\ln{n}}[/math]
[math]p_n\#\sim{n^n}[/math]
so we end up with
[math]\frac{n}{n^{n/n}}[/math]
Take note that [math]p_n\#>n[/math]
Let [math]\epsilon[/math]<0
Therefore we have:
[math]\frac{n}{n-\epsilon}\simeq{0}[/math]
>>7693329
You had the right idea, you just messed up through your work, though.
>>7693565
Huh? Why didn't those 2 LaTeX expressions not work? They worked in the TeX preview...
>>7693596
Can someone please make them work?
>>7693596
There's something wrong with the way 4chan parses the tex code, no one really knows what it is but putting white space everywhere seems to work.
>inb4 this doesn't work
[math] \ln{ ( p_n \#) } \sim{ p_n } \sim{n \ln{ n }} [/math]
[math] \frac{ n }{ n - \epsilon} \simeq{ 0 } [/math]
>>7693565
So does n/((n#)^(1/n)) converge to 0 or some positive value?
>>7694277
Alright so the sequence in the OP would be n/(e^(n*ln(n)))^(1/n).
This is a constant sequence, 1, which isn't the limit as seen from the picture.
Did I understand something wrong or is it just the approximate nature of these results?
Isn't it peculiar how the ratio between n and nth root of nth primorial converges? Isn't this kind of convergence just really nice?
Really? Don't you like this sort of convergence, /sci/? Why isn't anyone expressing their affection towards such an interesting sequence?
>>7695426
What kind of sequences do you like?
>>7695428
any sequence that illustrates the ideas I'm currently studying, serving as an example to aid in learning the theorems?
>>7693236
How to evaluate whether this is null or not?
>>7695433
gay
>>7696386
Is there no way to evaluate this?
>>7693236
Careful, you can't say slopes anymore. The proper term is "Asian Americans."
>>7697039
Thanks for bumping, it'd be really nice if someone knew how to evaluate whether or not the sequence is null or not.
Here's the values of the sequence at some 10^n:
n=10^1: 1.0445
n=10^2: 0.63573
n=10^3: 0.40472
n=10^4: 0.29179
n=10^5: 0.21880
Is this any indicator of being not null? 10^6 takes Maple too long to compute.
Wtf is a primorial? Is it some new hipster function?
Can convergence to 0 happen arbitrarily slowly so that the graph can appear indistinguishable from a non-null one when inspected at a "small" less than infinitely large interval?
Is it ever feasible to make statements about convergence based on the properties of the graph at finite intervals?
Is the sequence really that untouchable, /sci/?
>>7697336
[math]x\# = e^{\theta(x)}[/math], [math]\theta(x)[/math] being the first Chebyshev prime counting function.
Is it very probably null?
>>7693236
Stop bumping dude. It probably converges because the prime density converges as well.
>>7699461
*to a positive value
>>7699463
>to a positive value
Alright, that's great. Then doesn't the growth of the denominator have to converge to a constant? It can't keep curving more and more violently if the ratio converges to a positive value, right?
>>7699706
Does (n#)^(1/n) really asymptotically approach a line? If not, is it even possible for n/(n#)^(1/n) to not be null?
>>7693236
For large numbers:
ln(x#)=x
x#=e^x
(x#)^1/x = e
x/[(x#)^1/x] = x/e --> doesn't converge
it diverges, OP you are a retard if you can't even plot a function properly
