Prove that $\lim_{n\rightarrow\infty} \frac{n^3}{3^n}$ is finite.
How can I prove that this sequence converges? I tried so many ways but didn't succeed tried with the root test, integral test, ratio test
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maya cohen
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4What you wrote is not a series. The sequence converges to $0$ by L'Hopital. – lulu Apr 28 '22 at 20:41
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1Ratio test? Maybe you mean $\sum_{n\to \infty}n^3/3^n$ – Arnaldo Apr 28 '22 at 20:43
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1Intuitively:$$\dfrac{n^3}{3^n} = \underbrace{\dfrac{n}{3}\dfrac{n}{3}\dfrac{n}{3}\dfrac{1}{3}\dfrac{1}{3}\dfrac{1}{3}\dfrac{1}{3}\cdots \dfrac{1}{3}\dfrac{1}{3} }_{n \ \ fractions}\ < \ ???$$ – David P Apr 28 '22 at 20:51
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1By the ratio test $\sum n^\alpha x^n$ has same behaviour than $\sum x^n$ simply because $\frac {n+1}{n}\to 1$. So you just need to know the result for geometric series. – zwim Apr 28 '22 at 20:53
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1If you just want to find the limit, you can look at $\ln(n^3/3^n)$. Indication $(\ln x)/x \to 0$ as $x \to +\infty$. – Christophe Leuridan Apr 28 '22 at 21:22
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i know that 3^n is groing stronger but how do I write it ? – maya cohen Apr 28 '22 at 22:50
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1What you have is a sequence and not a series. – Gary Apr 28 '22 at 22:50
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I changed the word to sequence can you help me with this question please? – maya cohen Apr 28 '22 at 23:00
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2Prove by induction that $3^n > n^4$ for $n\geq 8$. Then $\frac{{n^3 }}{{3^n }} < \frac{1}{n}$ for $n\geq 8$. – Gary Apr 28 '22 at 23:28
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1Now that you've changed it to sequence, it's a duplicate of this question (and many others). – Hans Lundmark Apr 29 '22 at 05:20
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thanks ! so much sir – maya cohen Apr 29 '22 at 06:55
1 Answers
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Another approach :
Consider the series $\sum_{n\in\Bbb{N}}\frac{n^3}{3^n}$
Using Ratio Test,
$\frac{x_{n+1}}{x_n}=\frac{(n+1)^3}{3^{n+1}}×\frac{3^n}{n^3}=\frac{1}{3}(1+\frac{1}{n})^3$
$\lim_{n\to\infty} \frac{x_{n+1}}{x_n}= \frac{1}{3}<1$
Hence, $\sum_{n\in\Bbb{N}}\frac{n^3}{3^n}<\infty$ implies $\frac{n^3}{3^n}\to 0$
Sourav Ghosh
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1Please do not answer questions that do not meet the quality standards (see in particular "Spread the Word"). Also, this question is a duplicate. – Gary Apr 29 '22 at 07:15
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1@Gary Should I delete this?It would be a long comment. That's why I have included this as answer. – Sourav Ghosh Apr 29 '22 at 07:20
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hey , can you also help me , how to calculate the number that the sum is converging to ? – maya cohen Apr 29 '22 at 08:04
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1See https://math.stackexchange.com/questions/1598928/summing-sum-k-0-infty-frack33k?noredirect=1 – Gary Apr 29 '22 at 08:45