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Pamela

How do you solve this limit? 2017-02-15 06:15:24 Post No. 18052174

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How do you solve this limit? 2017-02-15 06:15:24 Post No. 18052174

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I know it diverges, but I don't find a way to prove it. I've tried using ln on both sides but i get to a plece where I get 0*infinity and There's nothing left to wotk with.

Any idea?

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Try taking the derivative of it.

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>>18052174

Use Stirling approximation

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L'Hopital rule i believe should work

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>>18052174

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

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>>18052211

Are you sure you can take the limit out of the logarithm, m8?

I suggest being careful with such things.

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>>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

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>>18052371

fuck i am retarded and missed factorial, w8

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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.

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>>18052195

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

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>>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

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>>18053508

Yeah, I just showed you here:

>>18052616

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>>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

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>>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)

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>>18054893

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

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

In that proving that it diverges is super ez

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