$\{\frac{\log n}{n^r}\}$ is bounded for positive rational $r$.

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.