Given two positive real sequences $(U_n)$ and $(V_n)$ with $(V_n)>0$ and two fixed real numbers $a$ and $b$, is it true that if $\underset{n \rightarrow +\infty}{\lim \sup} \text{ }\frac{U_n}{V_n} \leq a$ and $\underset{n \rightarrow +\infty}{\lim \inf} \text{ }\frac{U_n}{V_n} \geq b$, then $U_n=\Theta\left ( V_n \right )$ (i.e. $\exists k_{1}>0 \exists k_{2}>0\;\exists n_{0}\;\forall n>n_{0} : {\displaystyle k_{1}\cdot V_n\leq U_n\leq k_{2}\cdot V_n}$) ?
I feel like it is but I'm unsure how to prove it
EDIT : It's not true, thanks drhab. The thing that I am trying to do is to study the growth of a nonnegative real sequence $(R_n)$ that verifies $R_n<=A\ln n + B$ for some $A,B>0$ and $\lim \inf \frac{R_n}{\ln n} \geq C$ for some $C \geq 0$. Is there any way I could compare asymptotically $R_n$ to $\ln n$ with this information ?
Let $V_{n}=1$.
Let $U_{n}=1$ if $n$ is odd and $U_{n}=\frac{1}{n}$ otherwise.
Then $0=\liminf\frac{U_{n}}{V_{n}}<\limsup\frac{U_{n}}{V_{n}}=1$
But $\forall n>n_{0}\left[k_{1}\leq U_{n}\right]$ implies that $k_{1}\leq0$.