TITLE: Find the limit of $\frac{(n+1)^\sqrt{n+1}}{n^\sqrt{n}}$.
QUESTION [10 upvotes]: Find $$\lim_{n\to \infty}\frac{(n+1)^\sqrt{n+1}}{n^\sqrt{n}}$$
First I tried by taking $\ln y_n=\ln \frac{(n+1)^\sqrt{n+1}}{n^\sqrt{n}}=\sqrt{n+1}\ln(n+1)-\sqrt{n}\ln(n),$ 
which dose not seems to take me anywhere. Then I tried to use squeeze theorem, since $\frac{(n+1)^\sqrt{n+1}}{n^\sqrt{n}}\geq 1$, but I need an upper bound now. It's for a while I am trying to come up with it but stuck. Can you help me please.

REPLY [3 votes]: Note that
$$
\begin{align}
1\le\frac{(n+1)^{\sqrt{n+1}}}{n^{\sqrt{n}}}
&=\left(1+\frac1n\right)^{\sqrt{n}}(n+1)^{\sqrt{n+1}-\sqrt{n}}\\
&=\left(1+\frac1n\right)^{\large\frac{n}{\sqrt{n}}}
(n+1)^{\large\frac1{\sqrt{n+1}+\sqrt{n}}}\\
&\le e^{\large\frac1{\sqrt{n}}}
\left(\sqrt{n+1}^{\frac1{\sqrt{n+1}}}\right)^2
\end{align}
$$
The Squeeze Theorem then says that
$$
\lim_{n\to\infty}\frac{(n+1)^{\sqrt{n+1}}}{n^{\sqrt{n}}}=1
$$