$\lim \limits_{n \to \infty\ } \sqrt[n]{\left|\frac {1}{n3^n}-\frac {n^{170}}{4^n} \right|}= \ldots=\lim \limits_{n \to \infty\ } \sqrt[n]{\frac {1}{n3^n}} \cdot\sqrt[n]{\left|1-\left(\frac {3}{4}\right)^n \cdot n^{171}\right|}= $ $=\lim \limits_{n \to \infty\ }\frac {1}{3} \cdot \sqrt[n]{\left|1-\left(\frac {3}{4}\right)^n \cdot n^{171}\right| }\ $
I know that $\forall{q>1, k\in \Bbb N} \ \lim \limits_{n \to \infty\ }\frac {n^k}{q^n}=0$, but I don't know how show that $ \lim \limits_{n \to \infty\ } \left(\frac {3}{4}\right)^n \cdot n^{171}=0$
Best Answer
By root test
$$\sqrt[n] {a_n}=\sqrt[n]{\left(\frac {3}{4}\right)^n\cdot n^{171}}=\frac34 (\sqrt[n]n)^{171} \to \frac34 <1 \implies a_n \to 0$$
recall indeed that $\sqrt[n]n\to 1$.
As an alternative
$$\left(\frac {3}{4}\right)^n \cdot n^{171}=e^{\log\left(\left(\frac {3}{4}\right)^n \cdot n^{171}\right)}\to 0$$
indeed
$$\log\left(\left(\frac {3}{4}\right)^n \cdot n^{171}\right)=n\log\frac34+171\log n=-n\left(\log\frac43+171\frac{\log n}n\right)\to -\infty$$