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Prove that $$\lim_{n\rightarrow \infty}\sqrt[n]{\ln(n+1)} = 1$$

I already know that $\lim_{n\rightarrow \infty}\sqrt[n]{n} = 1$, but now sure how to use it.

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    $\begingroup$ Observe that $\log$ is a continuous function and hence the limit sign can be taken inside the log. $\endgroup$ – Foobaz John Jan 11 '16 at 22:01
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    $\begingroup$ If you know the behaviour of $\sqrt[n]{n}$, use $1\leqslant\ln(n+1)\leqslant n$ hence $$1\leqslant\sqrt[n]{\ln(n+1)}\leqslant\sqrt[n]{n}.$$ $\endgroup$ – Did Jan 11 '16 at 22:18
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Hint : $1\le\ln(n+1) \le n$ then use what you know.

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Use the theorem that when $a_n\to L$ then $\sqrt[n]{a_1\cdots a_n} \to L$ with $a_n=\frac{\ln (n+1)}{\ln n}$.

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Let $f(n) = \sqrt[n]{\ln(n+1)}$, and let $g(n) = \ln f(n) = \frac{1}{n}\ln(\ln(n+1))$. By L'Hôpital, we have $$\lim_{n\to\infty} g(n) = \lim_{n\to\infty}\frac{1}{\ln(n+1)}\frac{1}{n+1}=0.$$ Thus $$\lim_{n\to\infty} f(n) = 1$$ by continuity of $\ln$.

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Let y=$\sqrt[n]{\ln(n+1)}$

Take the natural log of both sides, $\implies ln(y)=\frac{ln(ln(n+1))}{n}$ Now either you can apply L.H rule, or simply see that n increases much faster than ln(ln(n+1)), So for n$\rightarrow \infty$ ln(y)=0, yielding y=1.

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