How do I evaluate the limit of things involving factorials?
Heres an example problem my textbook dismissed as "trivial". Am I missing something here?
try the ratio test and see what cancels
Factorials grow faster than just about everything
>>7965630
If you're looking for a formal answer, take the limit of e^log(10^n/n!) and simplify. Remember properties of logs, e.g. log(ab)=log(a)+log(b)
>>7965636
Actually, the exponential function grows faster than factorials.
So, 10^n > n!
Thus, the limit tends to infinity
>>7965642
you sure about that?
>>7965647
Fascinating.
But how come?
>>7965642
Factorial is faster bruh
>>7965654
My goodness, you're right. I see it now.
Thanks for the explanation.
OP here
>>7965637
This looks like what I'm looking for, but I have immediately hit another brick wall.
I can see intuitively why it works, but I am not sure how to formally show it.
Think about it like this.
[math]\frac{10^{n+1}}{(n+1)!}=\frac{10^{n}}{(n)!}\frac{10}{n+1}[/math]
It should be fairly obvious that when n is 10 or larger [math]\frac{10^{n}}{(n)!}[/math] is decreasing at a rapid pace.
>>7965642
Nope. Think of it like this: when you get to bigger numbers, how much is each of them increasing? If you go from n=99 to n=100, 10^n increases by a factor of 10 while n! increases by a factor of 100 (and only gets bigger).
>>7965651
e^log(10^n/n!) = e^(log(10^n)-log(n!))
= e^(nlog(10)-Sum_i=1^n(log(i)))
= e^(n-Sum(log(i)))
which goes to e^(-inf) for large n.
>>7965642
> mfw f(x)=log(x!) is concave up
>>7965630
Notice that the limit as n tends to 0 is trivial, since n! > 0 we can just evaluate at the limit point, the limit as n tends to infinity is a little trickier but you can intuitively think of it as at some point 10^n < n! Which means that at some point we can just consider the limit as 1/n!, which I'm hoping you can see tends to 0, more formally we could bound your function and then apply the squeeze theorem.
>>7965648
[eqn]\lim_{n\to\infty} \frac{10^{n}}{n!} = \lim_{n\to\infty} \frac{10}{n} \frac{10}{n-1} \frac{10}{n-2}...\frac{10}{k}...\frac{10}{2} \frac{10}{1}[/eqn]
>not using the ratio test.
Why you faggots lurk this board if you have the same skills as the local burger flipper?
>>7967644
You bumped this thread to post that?
>>7965648
from n=10 and on, factorial starts growing faster. at n=25, n!>10^n
>>7965630
Compare it to a geometric series (10/11)^n
eventually that series will grow faster than 10^n/n!
>>7965642
This implies that the taylor series for exponential functions have a radius of convergence of only one.
>>7965630
ye
here's a rule of thumb for ya:
log(n) << n << n^k << a^n << n! << n^n
where k >1 and a >1
to put it simply, factorials grow faster than exponentials
this is something you're allowed to just remember, no need for a test
>>7965630
look up
Stirling
>>7965630
Hint:
For n>11
[math]\frac{10^n}{n!}<\frac{10^{10}}{10!}\frac{10^n}{11^n}[/math]
>>7968723
Sry, the exponent on RHS should be n-11 not n
>>7965683
Holy shit
