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Differential Calculus

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9 Mar 2010 16:25:32 IST

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what is limit n tends to infinity (n!)^1/n ?

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what is limit n tends to infinity (n!)^1/n ?

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edison

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9 Mar 2010 19:31:08 IST

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ITS 1

as Lim_{n --> inf} n^1/n = 1

Thus Limit (n !)^1/n

= (1 . 2 . 3 .....n)^1/n

= 1^1/n .2^{^1/n.} ....n^1/n= 1.1.1....1 (n times) = 1

ThIrStY of IIT

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12 Mar 2010 09:46:24 IST

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i think it is 1 .is it right.

Hari Shankar

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14 Mar 2010 08:01:59 IST

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Nope. Thats badly off

By Stirling's formula $n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$ for large n

So $a_n = (n!)^{\frac{1}{n}}$ diverges

Mr.Nasty ® Retired from Action.

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14 Mar 2010 08:42:12 IST

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Formula is correct but for large value of n

1/n tends to zero

and we will raise Stirling's Formula to the power tending to zero.

That gives under root (2 pi n) ^ 1/n x n/e

Please clarify further.

Edited:

Thanks Hari Sir

Hari Shankar

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14 Mar 2010 10:00:10 IST

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I dont see how

$(n!)^{\frac{1}{n}} \approx \sqrt{2 \pi n}^{\frac{1}{n}} \ \frac{n}{e}$

and this obviously goes to infinity.

In fact you may be familiar with this limit $\lim_{n \rightarrow \infty} \frac{a^n}{n!} = 0$

This tells us that for any +ve real number a, we can find a natural number N such that N! > a^N

In other words given any real a, we have $(N!)^{\frac{1}{N}} > a$

That means the sequence $a_n = (n!)^{\frac{1}{n}}$ can be made arbitrarily large. This is just another way of

saying that $\lim_{n \rightarrow \infty} (n!)^{\frac{1}{n}} = \infty$ (in extended real numbers)

Vivek Roy

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14 Mar 2010 12:13:53 IST

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@Hari sir, while i was preparing for regional math olmpiad I came across this problem and indeed it is a diverging series....The answer is Infinity. One can check out www.wolframalpha.comhttp://www.wolframalpha.com/input/?i=limit+n--%3Einfinity+(n!)^1/n

Hari Shankar

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14 Mar 2010 14:31:43 IST

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good then. Only its the divergence of the sequence not the series. series usually refers to a sum of terms

sanjeev

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23 Sep 2011 19:42:16 IST

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the sequence with nth term (n!)^(1/n) is (1,1.414,1.817,2.2136,2.60,2.99,3.38........................) which appears to be unbounded above and hence sequence is not convergent....actually it diverges to + infinity

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