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