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I am improving my skill at formal sequence convergence proofs, I find them very tricky. I want to prove that:

$$\frac{n^5}{3^n} \rightarrow0$$

This should be read as "converges to zero", the question is, how large should $n$ be?

I would want a way to compare these two expressions I have trouble picking a big $n$ because I do not quite understand how to compare denominator and enumerator. Could someone drop a small hint so I can continue with my proof, I do not want to use any limit theorems. What I want is that for $n>n_0$:

$$\left| \frac{n^5}{3^n} \right|< \epsilon$$

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    "converges for large $n$" has little meaning. The truth is that it converges. –  Nov 16 '18 at 17:44
  • Ratio Test leads to a nice proof. https://en.wikipedia.org/wiki/Ratio_test – irchans Nov 16 '18 at 17:45
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    "the question is, how large should n be?" That depends on how close you want $\frac {n^5}{3^n}$ to be to $0$. If $n = 3$ then $\frac {n^5}{3^n} =9$. Annd $9$ is close enough to $0$ if you are shooting an arrow at the side of a barn from a distance of three feet. – fleablood Nov 16 '18 at 17:48
  • that made me laugh :) –  Nov 16 '18 at 17:49

5 Answers5

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Note that for $n \geqslant 6$, by binomial theorem, $$ 3^n = (1+2)^n = \cdots \geqslant ? $$

xbh
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You'd like $\dfrac{n^5}{3^n} < \epsilon$.

This is equivalent to $5 \log n - n \log 3 < \log \epsilon$.

Can you show that $5 \log n - n \log 3 \to -\infty$?

Umberto P.
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What you need to show is that for any $\varepsilon > 0$, there exists an $N$ so large that $n > N$ implies $n^5/3^n < \varepsilon$. Alternatively you can use l'hopitals rule.

Daniel Xiang
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Consider the ratio of two successive terms,

$$\frac{t_{n+1}}{t_n}=\frac13\left(1+\frac1n\right)^5.$$

When

$$1+\frac1n<\sqrt[5]3$$ i.e. $n>4$, it becomes less than one, and from there the sequence decreases to $0$.

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Let $e^y =3$ where $y >1.$

Then $\dfrac{n^5}{e^{(yn)}} \lt (\dfrac{6!}{y^6})\cdot \dfrac{1}{n}.$

Peter Szilas
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