Find an equivalent sequence

Question

Consider $ u_n = (n+1)^{1/n+1} - n^{1/n} $

Find an equivalent sequence at infinity. (meaning $ u_n / y_n \rightarrow 1 ) $

I tried doing :

$ u_n = e^{ \frac{ln(n+1)}{n+1}}(1 - e^{\frac{ln(n)}{n} - \frac{ln(n+1)}{n+1} } ) $

Then $ \frac{ln(n)}{n} - \frac{ln(n+1)}{n+1} = \frac{ -1 + o(1) + ln(n)}{n(n+1)} $

How do I get a simple equivalent? Thanks in advance.

Answer 1

We have \begin{align} u_n &= \exp(\frac{\ln(1+n)}{1+n}) - \exp(\frac{\ln(n)}{n})\\ &= \exp(\frac{\ln(1+n)}{1+n})(1 - \exp(\frac{\ln(n)}{n}-\frac{\ln(1+n)}{1+n}))\\ \end{align} With $$ 1 - \exp(\frac{\ln(n)}{n}-\frac{\ln(1+n)}{1+n}) \sim \frac{\ln(1+n)}{1+n} - \frac{\ln(n)}{n} $$ and $$ \exp(\frac{\ln(1+n)}{1+n}) \sim 1 $$ we get $$ u_n \sim \frac{\ln(1+n)}{1+n} - \frac{\ln(n)}{n}. $$ Also, you can develop further and see that \begin{align} \frac{\ln(1+n)}{1+n} - \frac{\ln(n)}{n} \sim \frac{1-\ln(n)}{n^2}. \end{align}

Answer 2

$$(n+1)^{1/(n+1)}-n^{1/n}=\sqrt[n]{n}\left(\frac{(n+1)^{1/(n+1)}}{n^{1/n}}-1\right)$$ Take $v_n=\left(\frac{(n+1)^{1/(n+1)}}{n^{1/n}}-1\right)$. Then $\frac{u_n}{v_n}=\sqrt[n]{n}\to1$.

Answer 3

Try $y_n=n$. We have $$\begin{align} \lim_{n\to\infty}{u_n\over y_n}&=\lim_{n\to\infty}{(n+1)^{1/n+1} - n^{1/n}\over n}\\&=\lim_{n\to\infty}{n+1\over n}(n+1)^{\frac1n}-{n^{1/n}\over n}\\ &= 1\cdot 1-0\\&=1 \end{align}$$