$\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$

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

$$\left(\frac {3}{4}\right)^n \cdot n^{171}=e^{\log\left(\left(\frac {3}{4}\right)^n \cdot n^{171}\right)}\to 0$$


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

  • $\begingroup$ I can't use methods that haven't been proven at my lectures. $\endgroup$ – matematiccc Nov 15 '18 at 20:19
  • $\begingroup$ @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 . $\endgroup$ – user Nov 15 '18 at 20:27
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    $\begingroup$ @gimusi The relevant term is $\log\left(1-(3/4)^n\, n^{171}\right)\sim -(3/4)^n\, n^{171}$ $\endgroup$ – Mark Viola Nov 16 '18 at 4:16
  • $\begingroup$ 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! $\endgroup$ – user Nov 16 '18 at 6:34

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