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I have found this following problem on sequence:

Show that for any positive rational number $r$, the sequence $\{\frac{\log n}{n^r}\}$ is bounded.

My Solution. let us consider $f(x)=\frac{\log x}{x^r}$. Then $~f'(x)=x^{-(1+r)}(1-r\log x)$. Then $f$ is monotone decreasing for $x>e^{1/r}$. And by derivative test we get it has a maximum value at $x=e^{1/r}$ (since, $f"(e^{1/r})<0$) Therefore, the given sequence is bounded above and also each term of the sequence is nonnegative so its bounded below too. Therefore the sequence is bounded.

But I cannot understand why $~r$ is given rational? I mean I don't use it to be rational ......I just use that $r$ is positive real.

I am confused here....!! please help me to figure out my mistakes if there be any ...!! Thank you.

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  • $\begingroup$ $r$ need not be positive rational as you observed. $\endgroup$ – Hanul Jeon Apr 30 '18 at 4:47
  • $\begingroup$ Thank you....!! $\endgroup$ – Indrajit Ghosh Apr 30 '18 at 4:48
  • $\begingroup$ Hmm, I can only conclude the aren't assuming real exponents are defined yet and that want a more direct prove involving values. $\endgroup$ – fleablood Apr 30 '18 at 4:52
  • $\begingroup$ You don't have any mistakes. It is valid for all $r>0$. We can also observe that for $r>0$ we have $\sup_{n\in \Bbb N} (\log n)/n^r\leq $ $\sup_{x\geq 1}(\log x)/x^r=$ $(1/r)\sup_{x\geq 1}(\log (x^r))/x^r=$ $(1/r)\sup_{y\geq 1}(\log y)/y.$.... (because $\{x^r: x\geq 1\}=\{y:y\geq 1\})$....so it can also be solved by showing that $(\log y)/y$ is bounded on $ y\in[1,\infty).$ $\endgroup$ – DanielWainfleet Apr 30 '18 at 6:13
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Your observation is corrcet. Your sequence is bounded if $r$ is positive real, which is weaker than you mentioned.

Let me introduce another proof of your problem: we know that $\lim_{n\to\infty} \log n / n^r=0$ when $r$ is positive real number. Boundedness of our sequence easily follows from the fact that every convergent sequence is bounded.

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  • $\begingroup$ The fact that this becomes a null sequence is : This is m.d. and bounded below by infimum '0'....Am I correct..? $\endgroup$ – Indrajit Ghosh Apr 30 '18 at 4:53
  • $\begingroup$ @IndrajitGhosh I can't guess the meaning of m.d.. Could you explain it what it means? $\endgroup$ – Hanul Jeon Apr 30 '18 at 4:58
  • $\begingroup$ I meant "Monotone Decreasing".. $\endgroup$ – Indrajit Ghosh Apr 30 '18 at 4:58
  • $\begingroup$ @IndrajitGhosh These two facts does not explain it converges to 0, though they implies the boundedness of the sequence. $\endgroup$ – Hanul Jeon Apr 30 '18 at 5:03
  • $\begingroup$ ...this is a true fact...see en.wikipedia.org/wiki/Monotone_convergence_theorem $\endgroup$ – Indrajit Ghosh Apr 30 '18 at 5:07
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Another way to solve would be to use L’Hospital’s rule for the limit $$lim_{x \to \infty} \frac {log x}{x^{r}}$$. From here it easily follows that it is bounded by zero for every real positive r. I believe that powers to irrationals may not be well defined in your course.

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The restriction $r\in\mathbb{Q} ^{+} $ probably means that the problem has to be solved using basic algebra and not using any calculus based methods.

Use the fundamental inequality $$\log x\leq x-1, x>0$$ with $x=n^r$ to get $$r\log n\leq n^r-1<n^r$$ and hence $$\frac{\log n} {n^r} <\frac{1}{r}$$ and you are done.

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  • $\begingroup$ Well...It seems that your solution don't need the restriction on $r$...Isn't it? $\endgroup$ – Indrajit Ghosh May 1 '18 at 7:42
  • $\begingroup$ @IndrajitGhosh: the restriction on $r$ is used for the definition of symbol $n^r$ and the identity $\log n^r=r\log n$. If $r$ is irrational then the proof of $\log n^r=r\log n$ is not algebraic. Perhaps I should have stated this explicitly. $\endgroup$ – Paramanand Singh May 1 '18 at 15:34

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