but then you can reason that n! as n approaches infinity is infinity multiplied to itself an infinite amount of times, or infinity^infinity, and (infinity^infinity)^(1/infinity) = infinity^1 = infinity
Yes things get tricky when one tries to do arithmetic with infinities; not all of the usual rules apply. This is part of the reason for introducing limits.
∞^0 is one of the indeterminate forms. An indeterminate form is a form an expression can take when evaluating a limit. If you get an indeterminate form, it means that you need to do more work to determine the limit. In the case of ∞^0, it's possible that the infinity represents a growing expression, while the 0 represents an expression approaching 0. When this happens, it's possible for the ∞ part to "overpower" the 0 part. One example is the case shown in the video. A simpler example is the limit of (n^n)^(1 / n) as n approaches infinity. If you just plug in n = ∞ and work things out, you'll end up with ∞^0. But you can instead simplify the original expression first: (n^n)^(1 / n) = n^(n * 1 / n) = n^(n / n) = n^1 = n. Now you can take the limit as n approaches infinity. As you can see, the limit is infinity, not 1.
[edit] The other limit of (n!)^(1/n) where n -> 0: exp((ln(x!)/x) -> numerator and denominator are 0 l'Hopital's Rule d/dx ln(x!))/x = digamma(x+1) * gamma(x+1) / 1 plug in x and the solution in e^digamma(1) = e^(-gamma) =~ 0.56146
i used the stirling approximation and for large n it converges to (n/e), so the limit (n->inf) -> inf
nice solution bro
Thanks!
i got infinity raised to 0, as 1/n approached 0 and n! approached infinity, which would then be 1, right?
but then you can reason that n! as n approaches infinity is infinity multiplied to itself an infinite amount of times, or infinity^infinity, and (infinity^infinity)^(1/infinity) = infinity^1 = infinity
Yes things get tricky when one tries to do arithmetic with infinities; not all of the usual rules apply. This is part of the reason for introducing limits.
∞^0 is one of the indeterminate forms. An indeterminate form is a form an expression can take when evaluating a limit. If you get an indeterminate form, it means that you need to do more work to determine the limit.
In the case of ∞^0, it's possible that the infinity represents a growing expression, while the 0 represents an expression approaching 0. When this happens, it's possible for the ∞ part to "overpower" the 0 part. One example is the case shown in the video.
A simpler example is the limit of (n^n)^(1 / n) as n approaches infinity. If you just plug in n = ∞ and work things out, you'll end up with ∞^0. But you can instead simplify the original expression first:
(n^n)^(1 / n)
= n^(n * 1 / n)
= n^(n / n)
= n^1
= n.
Now you can take the limit as n approaches infinity. As you can see, the limit is infinity, not 1.
[edit] The other limit of (n!)^(1/n) where n -> 0:
exp((ln(x!)/x) -> numerator and denominator are 0
l'Hopital's Rule
d/dx ln(x!))/x = digamma(x+1) * gamma(x+1) / 1
plug in x and the solution in e^digamma(1) = e^(-gamma) =~ 0.56146