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Calculate:

$$\displaystyle{\lim_{n \to \infty}} \sqrt[n]{(\ln(n))^{2009}-1}$$

I thought about some trick like this one:

$$\displaystyle{\lim_{n \to \infty}} \sqrt[n]{(\ln(n))^{2009}-1} = \displaystyle{\lim_{n \to \infty}} \sqrt[n]{\left(\frac{e^{2009*\ln[\ln(n)]}-1}{2009*\ln[\ln(n)]}\right) * 2009*\ln[\ln(n)]}$$

But unfortunately it doesn't seem to lead anywhere.

It is important also to add that I can not use L-Hospitals rule.

Falcon
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theboyboy
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2 Answers2

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For $n\geq 9$, we have $\ln(n)>2$. Also, $\ln(n)<n$. Hence, $$ 1<\left(\ln(n)\right)^{2009}-1<\left(\ln(n)\right)^{2009}<n^{2009}\\ \Rightarrow1<\sqrt[n]{\left(\ln(n)\right)^{2009}-1}<n^{2009/n}=\left(n^{1/n}\right)^{2009} $$ Using $\lim_{n\to\infty}n^{1/n}=1$ and the squeeze theorem, it follows that your limit is $1$. See here, for instance, about proofs of this limit.

bjorn93
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$\ln n=t$

take log on both side $$\log L=\lim_{t\to \infty} \frac{{(t)}^{2009}-1}{e^t}$$

By taylor series$$=\lim \frac{1-\frac{1}{t^{2009}}}{\frac{1}{t^{2009}}+\frac{t}{t^{2009}}...+\frac{t^{2010}}{2010!t^{2009}}+....}=0$$

$$L=e^0=1$$

Albus Dumbledore
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    And without L-Hospitals rule? I can't use L-Hospitals rule, I'm sorry, I will edit my question to include that information. Nevertheless, thank you for your help. – theboyboy Nov 28 '20 at 03:59