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It is obvious that the limit is $0$, but, how can we prove that?
What I did: We can see that $n^3$ monotonously grows to infinity, thus $$\lim_{t \rightarrow \infty}\frac{c}{t}=0, \text{for some fixed } c$$

FisherDisinformation
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A6SE
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  • Basically yes, while your notation should use limits and not "infinity" in the denominator – FisherDisinformation Aug 12 '15 at 13:03
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    You can prove this by showing for any positive $\epsilon$ that there is some $n$ such that $\frac{5}{n^3} < \epsilon$. Because the function montonically decreases, this is enough to show that it limits to $0$. – Colm Bhandal Aug 12 '15 at 13:03
  • of course there are several ways to do that, but for starters just apply the definition of convergence. – user190080 Aug 12 '15 at 13:12
  • yes, so basically you pick something like $n > \sqrt[3]{\frac{5}{\varepsilon}}$, so that later you bound your sequence with $|\frac{5}{n^3}-0|<\varepsilon$ – FisherDisinformation Aug 12 '15 at 13:12

3 Answers3

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For all $n > 1$, $$0 < \frac{5}{n^3} < \frac 5n$$

Now, assuming you've already established that $\displaystyle \frac 1n \to 0$, apply the Sandwich Theorem (aka Squeeze Lemma, ...).

Simon S
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Let $\epsilon \gt 0$ be given. Then $ \exists n_0 \in \Bbb N$ such that $\frac 5\epsilon \lt n_0^3$ by Archimedean property.

Then for $n \gt n_0$ we get $\frac 1{n^3} \lt \frac 1{n_0^3}$ $\Rightarrow \frac 5{n^3} \lt \frac 5{n_0^3} \lt \epsilon$.

$\Rightarrow |\frac 5{n^3}| \lt |\frac 5{n_0^3}| \lt \epsilon$ $\Rightarrow |\frac 5{n^3} - 0| \lt \epsilon$.

Thus $\lim \frac 5{n^3} = 0$.

Error 404
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  • If I understood correctly, you approximated $\frac{5}{n_0^3}$ to be $0$, since as $n$ grows it gets much bigger than $n_0$? – A6SE Aug 12 '15 at 13:26
  • @A6Tech I didn't approximate it. See $|\frac 5{n^3}| \lt |\frac 5{n_0^3}| \lt \epsilon$. So I can simply 'ditch' the middle term i.e. $|\frac 5{n_0^3}|$ to get $|\frac 5{n^3}| \lt \epsilon$ and $|\frac 5{n^3}|=|\frac 5{n^3} - 0|$. – Error 404 Aug 13 '15 at 05:51
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Start by assuming that $\exists L>0$ such that L is the limit of the sequence $a_n$. In such case, $\forall n>n_0 \ a_n$ is within an $\epsilon-$limit of L. Let's say $\epsilon = 0.01$. Now use the monotonicity of the series and show that $\frac{c}{(n+1)^3}$ is outside of this interval. Hence the only true value is $L=0$.

Alex
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