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How to calculate the limit of $(n+1)^{\frac{1}{n}}$ as $n\to\infty$?

I know how to prove that $n^{\frac{1}{n}}\to 1$ and $n^{\frac{1}{n}}<(n+1)^{\frac{1}{n}}$. What is the other inequality that might solve the problem?

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  • $\begingroup$ Why not use the value of $\lim \bigl[(n+1)/n\bigr]^{1/n}$? $\endgroup$ – Lubin Sep 5 '12 at 0:26
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With $$y=\lim_{n\to\infty} (n+1)^{1/n},$$ consider, using continuity of $\ln$, $$\ln y=\lim_{n\to\infty} \frac{1}{n}\ln(n+1)=0.$$ This tells you that your limit is $1$.

Alternately, $$n^{1/n}<n^{1/n}\left(1+\frac{1}{n}\right)^{1/n}<n^{1/n}\left(1+\frac{1}{n}\right),$$ where the middle guy is your expression.

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  • $\begingroup$ I would prefer a proof that not involves other concepts but convergence of know sequences. $\endgroup$ – Jön Sep 4 '12 at 22:34
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For the other inequality, you could use $$ (n+1)^{\frac1n}\leq (2n)^{\frac1n}=2^{\frac1n}\,n^{\frac1n}. $$

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What about $n^{1/n}\lt (n+1)^{1/n}\le (2n)^{1/n}=2^{1/n}n^{1/n}$, then squeezing.

Or else, for $n \ge 2$, $$n^{1/n}\lt (n+1)^{1/n}\lt (n^2)^{1/n}=(n^{1/n})(n^{1/n}).$$ Then we don't have to worry about $2^{1/n}$.

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