How do you solve this limit?

Try taking the derivative of it.

>>18052174
Use Stirling approximation

L'Hopital rule i believe should work

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>>18052174

You should be able to solve it from here, friend!

>>18052211
Are you sure you can take the limit out of the logarithm, m8?
I suggest being careful with such things.

>>18052226
Yes, log is a monotonically increasing function. It's the same reason why you can take the limit out of e.

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lel m8

>>18052371
fuck i am retarded and missed factorial, w8

Note that (n/2)^(n/2) < n!

So taking the nth root of both sides gives:

Sqrt (n/2) < n!^(1/n)

Take the limit of both sides as n goes to inf and you see that the left side goes to inf. Therefore the right side must be inf as well.

>>18052195
That's a way to do it, but isn't there something more simple?

>>18052211
We can take n! as n(n-1)... And then have ln(n)+ln(n-1)... But there's nothing left to do because we can't distribute the limit if there's something wrong and each term gives infinity over infinity

>>18053508
Yeah, I just showed you here:
>>18052616

>>18053967
Yeah, yours seems like the most easy approach but how do you know that the inequality that you proposed holds true? It isn't that evident

>>18054636
n!=n*(n-1)*(n-2)*....*(n/2)*...*2*1
>n*(n-1)*(n-2)*....*(n/2)
>(n/2)*(n/2)*....*(n/2) = (n/2)^(n/2)

>>18054893
Shit those are supposed to be greater than signs, not greentext

Wait, isn't the nth root of n! just (n-1)!

In that proving that it diverges is super ez