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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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    Why not use the value of $\lim \bigl[(n+1)/n\bigr]^{1/n}$?2012-09-05

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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} where the middle guy is your expression.

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    I would prefer a proof that not involves other concepts but convergence of know sequences.2012-09-04
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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 $2^{1/n}$.