<s>[INST] Sequence $(u_n)$: $\left\{\begin{array} .u_1=1\\ u_{n+1}=\frac{u_n}{2^nu_n+3} \quad , n=1,2,3,\dots \end{array}\right.$ The sequence $(u_n)$ is defined by the formula:
$$\left\{\begin{array}
.u_1=1\\
u_{n+1}=\frac{u_n}{2^nu_n+3} \quad , n=1,2,3,\dots
\end{array}\right.$$
Find $\lim\limits_{n\to \infty}\sqrt[n]{u_n}$.
The first, I see:
$$u_{n+1}=\frac{u_n}{2^nu_n+3}<u_n\Rightarrow\sqrt[n+1]{u_{n+1}}<\sqrt[n+1]{u_n}<\sqrt[n]{u_n}$$
But, I forgot that $0<u_n<1$ so $\sqrt[n+1]{u_n}>\sqrt[n]{u_n}$.
If I can prove that the $(\sqrt[n]{u_n})$ is a decrease/increase sequence, it's easy to see $\lim\limits_{n\to \infty}\sqrt[n]{u_n}=\frac{1}{2}$.
Please help me. Thank you.
 [/INST] If you don't insist on using a monotonicity argument, you can also derive an explicit form for $u_n$ and then calculate the limit:
Setting $v_n := \frac{1}{u_n}$ for all $n$, we get
\begin{alignedat}{2}
&&v_{n+1} &= 2^n + 3v_n\\
&\implies 
&\frac{v_{n+1}}{3^{n+1}}-\frac{v_{n}}{3^{n}} &= \frac{2^n}{3^{n+1}}\\
&\implies 
&\sum_{k=1}^{n-1} \left(\frac{v_{k+1}}{3^{k+1}}-\frac{v_{k}}{3^{k}}\right) &=  \frac{1}{3}\sum_{k=1}^{n-1} \left(\frac{2}{3}\right)^k = \frac{1}{3}\frac{\frac{2}{3} -\left(\frac{2}{3}\right)^{n}}{1-\frac{2}{3}}\\
&\implies
& \frac{v_{n}}{3^{n}} - \frac{v_{1}}{3} &= \frac{2}{3} -\left(\frac{2}{3}\right)^{n}\\
&\implies
&v_{n} &= 3^n - 2^n\\
&\implies
&\sqrt[n]{u_n} &= \frac{1}{\sqrt[n]{3^n-2^n}} \xrightarrow{ n \to \infty } \frac{1}{3}.
\end{alignedat}
</s>