How to calculate $\lim\limits_{n\to \infty} \frac{(n+3)^{\sqrt{n+1}}}{(n+1)^{\sqrt{n}}}$

Question

I have no idea how to find a limit of sequence $$a_n= \frac{(n+3)^{\sqrt{n+1}}}{(n+1)^{\sqrt{n}}}$$. I think that it can be easily bounded by 1 from down, but this shows only $1≤a_n$. I think also it can be shown quite easily that we need only to show that limit of a sequence $n^{\sqrt{n+1}-\sqrt{n}}$ is equal to $1$, I am unable to show that too.

Answer 1

shows that using conjugation that \begin{split} a_n=\frac{(n+3)^{\sqrt{n+1}}}{(n+1)^{\sqrt{n}}} &= &\exp \left( {\sqrt{n+1}}\ln(n+3)- {\sqrt{n}}\ln(n+1)\right)\\ &=& \exp \left( \ln(n+3)\color{red}{\left(\sqrt{n+1}-\sqrt{n}\right)}+\sqrt{n}[\color{blue}{\ln(n+3)-\ln(n+1)} ]\right)\\ &=& \exp \left( \frac{\ln(n+3)}{\color{red}{\sqrt{n+1}+\sqrt{n}}}+\sqrt{n}\color{blue}{\ln\frac{n+3}{n+1}}\right)\\ \end{split}

On the other hand, we have $$ \lim_{n\to \infty}\sqrt{n}\ln\frac{n+3}{n+1} = \lim_{n\to \infty}\frac{2\sqrt{n}}{n+1}\times\frac{\ln\left(1+\frac{2}{n+1}\right)}{\frac{2}{n+1}} = 0\times1= 0$$ Given that, $\frac{\ln(x+1)}{x}\to 0$ as $x\to0$ whereas $$\lim_{n\to \infty}\frac{\ln(n+3)}{\sqrt{(n+1)}+\sqrt{n}} =\lim_{n\to \infty} 2\frac{\ln\sqrt{n+3}}{\sqrt{n+3}}\times \frac{\sqrt{1+\frac{3}{n}}}{\sqrt{1+\frac{1}{n}}+1}= 2\times0\times \frac{1}{\sqrt{2}}=0 $$

Given that $$\lim_{x\to \infty} \frac{\ln x}{x} = 0$$ so $$\color{brown}{\lim_{n\to \infty}a_n =\lim_{n\to \infty} \frac{(n+3)^{\sqrt{n+1}}}{(n+1)^{\sqrt{n}}}=1}$$

Answer 2

Few of the steps/limits here require justification,I'll leave that to you,if you need more help feel free to comment. $$a_n=\frac{\frac{(n+3)^\sqrt{n+1}}{n^\sqrt{n}}}{(1+\frac1n)^\sqrt{n}}=\frac{\frac{(n+3)^\sqrt{n+1}}{n^{\sqrt{n}+\sqrt{n+1}-\sqrt{n+1}}}}{(1+\frac1n)^\sqrt{n}}=\frac{(1+\frac3n)^{\sqrt{n+1}}}{n^{\sqrt{n}-\sqrt{n+1}}(1+\frac1n)^\sqrt{n}}$$ Lets find $\lim_{n\to\infty}n^{\sqrt{n}-\sqrt{n+1}}$ first notice $$\sqrt{n}-\sqrt{n+1}=\frac{(\sqrt{n}-\sqrt{n+1})(\sqrt{n}+\sqrt{n+1})}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n}^2-\sqrt{n+1}^2}{\sqrt{n}+\sqrt{n+1}}=-\frac{1}{\sqrt{n}+\sqrt{n+1}}\\\lim_{n\to\infty}n^{\sqrt{n}-\sqrt{n+1}}=\lim_{n\to\infty}n^{-\frac{1}{\sqrt{n}+\sqrt{n+1}}}=\lim_{n\to\infty}e^{-\frac{\ln(n)}{\sqrt{n+1}+\sqrt{n}}}=e^{\lim_{n\to\infty}-\frac{\ln(n)}{\sqrt{n+1}+\sqrt{n}}}\\\lim_{n\to\infty}-\frac{\ln(n)}{\sqrt{n+1}+\sqrt{n}}=0$$ I rewrote $n^{-\frac{1}{\sqrt{n}+\sqrt{n+1}}}$ as $e^{\ln(n^{-\frac{1}{\sqrt{n}+\sqrt{n+1}}})}$ Also $$\lim_{n\to\infty}(1+\frac1n)^{\sqrt{n}}=\lim_{n\to\infty}((1+\frac1n)^n)^{1/\sqrt{n}}=1$$ Almost analogous it can be shown that $$\lim_{n\to\infty}(1+\frac3n)^{\sqrt{n+1}}=1$$ Combining that together we get that $\lim a_n=1$