sequences and series - Find the limit of $frac{(n+1)^sqrt{n+1}}{n^sqrt{n}}$.

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.

Answer

Notice that for $f(x)=\sqrt x \ln x$ you have

$$f'(x)=\frac{\ln x}{2\sqrt x}-\frac1{\sqrt x}.$$ Now by mean value theorem $$f(n+1)-f(n) = f'(\theta_n)$$ for some $\theta_n$ such that $n<\theta_n

If we notice that $\lim\limits_{x\to\infty} f'(x) = 0$ we get that

$$\lim\limits_{n\to\infty} (\sqrt{n+1}\ln(n+1)-\sqrt{n}\ln n) = \lim\limits_{n\to\infty} f'(\theta_n) = 0.$$