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

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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)

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