An example to show that we can have $\limsup \frac{a_{n+1}}{a_n}=1$ and Cauchy root test $\limsup a_n^{\frac{1}{n}} <1$

Question

ok so I need to try and find a sequence satisfying $\limsup \frac{a_{n+1}}{a_n}=1$ and $\limsup a_n^{1/n}<1$.

I’ve tried lots of examples but most give both of them equal can someone just give me some help please?

Answer 1

Take the sequence$$1,1,\frac12,\frac12,\frac1{2^2},\frac1{2^2},\frac1{2^3},\frac1{2^3},\ldots$$In other words, $a_n=\dfrac1{2^{\left\lfloor\frac{n-1}2\right\rfloor}}$. Then$$\limsup_n\frac{a_{n+1}}{a_n}=1\text{ and }\limsup_n\sqrt[n]{a_n}=\frac1{\sqrt2}.$$

Answer 2

Take the two fold geometric series with two ratios $a$ and $b$,

$$a+ab+a^2b+a^2b^2+a^3b^2+a^3b^3+\cdots $$

Then $$\lim\limits_{n\to \infty}\sqrt[n]{a_n}=\sqrt{ab}$$

and $$\frac{a_{n+1}}{a_n}=\begin{cases} a&n\text{odd}\\ b&n\text{even}\\ \end{cases}$$

So $$\limsup\limits_{n\to \infty}\frac{a_{n+1}}{a_n}=\max{\{a,b\}}$$ $$\liminf\limits_{n\to \infty}\frac{a_{n+1}}{a_n}=\min{\{a,b\}}$$