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

Question

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?

Answer 1

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.

Answer 2

For the other inequality, you could use $$ (n+1)^{\frac1n}\leq (2n)^{\frac1n}=2^{\frac1n}\,n^{\frac1n}. $$

Answer 3

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 $2^{1/n}$.