# Evaluate $\lim \limits_{n \to \infty\ } \sqrt[n]{\left|\frac {1}{n3^n}-\frac {n^{170}}{4^n}\right|}$

$$\lim \limits_{n \to \infty\ } \sqrt[n]{\left|\frac {1}{n3^n}-\frac {n^{170}}{4^n} \right|}= \ldots=\lim \limits_{n \to \infty\ } \sqrt[n]{\frac {1}{n3^n}} \cdot\sqrt[n]{\left|1-\left(\frac {3}{4}\right)^n \cdot n^{171}\right|}=$$ $$=\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 $$\lim \limits_{n \to \infty\ } \left(\frac {3}{4}\right)^n \cdot n^{171}=0$$

• Take logarithm and factor. – hamam_Abdallah Nov 15 '18 at 20:12
• $n^{171}$ over $(4/3)^n$. Take $q=4/3$. – Sungjin Kim Nov 15 '18 at 20:24
• @i707107 That's it! – matematiccc Nov 15 '18 at 20:31

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

$$\log\left(\left(\frac {3}{4}\right)^n \cdot n^{171}\right)=n\log\frac34+171\log n=-n\left(\log\frac43+171\frac{\log n}n\right)\to -\infty$$

• I can't use methods that haven't been proven at my lectures. – matematiccc Nov 15 '18 at 20:19
• @matematiccc I've added an alternative, provided you know that $\log n/n \to 0$ which is a standard limit. Note that also $\frac{n^a}{b^n} \to 0 \quad b>1$ is also a standard limit . – user Nov 15 '18 at 20:27
• @gimusi The relevant term is $\log\left(1-(3/4)^n\, n^{171}\right)\sim -(3/4)^n\, n^{171}$ – Mark Viola Nov 16 '18 at 4:16
• Yes you are right! But the OP already had obtained that, the problem was to show that this term goes to zero. I choose one way for that! – user Nov 16 '18 at 6:34