I'm trying to evaluate $$\lim_{n \to +\infty}\frac {3^n}{n^3 \cdot 2^n}$$
Can someone help me? It should be 0
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Michael Rozenberg
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Anne
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No it isn't 0, it's $+\infty$. – user296113 Jan 22 '18 at 18:32
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It is $\frac{1.5^n}{n^3}$. – Math Lover Jan 22 '18 at 18:32
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$\sum_n \frac{n^3}{1.5^n} < \infty$ – mathworker21 Jan 22 '18 at 18:34
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1Possible duplicate of How to prove that exponential grows faster than polynomial? – Jan 22 '18 at 18:34
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Note that $$\frac {3^n}{n^3 \cdot 2^n}=e^{n\log 3-3\log n-n\log 2} \to +\infty$$ indeed $$n\log 3-3\log n-n\log 2=n\left(\log 3-\log 2-3\frac{\log n}{n}\right)\to +\infty$$ – user Jan 22 '18 at 22:28
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Prove that $$\left(\frac{3}{2}\right)^n>n^4$$ for all $n\geq40.$
Michael Rozenberg
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Yes, I did,$ \frac {3^n}{n^3}>2^n $indeed and the limit should be $\infty$, but the associated series should converge because I tried the ratio test and root test and it converges, – Anne Jan 22 '18 at 18:54
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The limit is equal to $+\infty$ because for $n\geq40$ we have $\frac{3^2}{n^32^n}\geq n\rightarrow+\infty$. – Michael Rozenberg Jan 22 '18 at 19:06
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