This is my study in Desmos.
Does $\Biggl(m\Bigl((k^\beta+1)^{1\over\beta}\Bigr)\Biggr)_{k\in\mathbb N}\to 1$ iff $\beta\gt2$?
It is clear that $m(k)=1,\forall k\in\mathbb N$. So the observation above seems weird, why shouldn't it be the same case $\forall\beta$?
Any help will be appreciate. Thank you!
In case someone will ask, the following (to the end of this post) is the jorney of how I get to this question
(I am afraid that reading following will make someone more confused.)
I was originally trying to give counter example to :
$(s_n)_{n\in\mathbb N}$ is a unbounded sequence, $\lim_n\frac{s_{n+1}}{s_n}=1$ $\implies\lim_ns_n=+\infty$.
I believe the statement is false because I believe I found another counter example, but I think this is quite interesting.
$(p(n))_{n\in\mathbb N}$, or more precisely, $(p_\beta(n))_{n\in\mathbb N}$, is the sequence $s_n$ I want to study.
Originally I take $\beta=2$, but then I realise that I can't prove $\lim_n\frac{s_{n+1}}{s_n}=1$, and then I start to consider subsequences.
If I take $n_k=k^\beta$, I expect $\lim_n\frac{s_{n_k+1}}{s_{n_k}}=1$. But the funtion $q(x)\to1$ (or equivalently, $h(x)\to1$) iff $\beta\gt2$, i.e. $\alpha\gt0$.