0
$\begingroup$

I would like to know if my proof is correct:

I want to show that the series:

$$a_n=\sum_{n=1}^{\infty}\frac{nx}{1+n^5x^2}$$

uniformly converges for $|x|\lt \infty$

I looked at $a_n\lt b_n=\sum_{n=1}^{\infty}\frac{nx}{n^5x^2}=\sum_{n=1}^{\infty}\frac{1}{n^4x}$

I used d'Alambert therorem: $\frac{a_{n+1}}{a_n}=\frac{n^4x}{(n+1)^4x}=(\frac{n}{n+1})^4\lt1$

so $a_n$ also converges for every x.

I'm not sure if showing that the series converges for for every x is enough since my book is not clear enough about it.

by the section in the book I think I was expected to use weierstrass' uniform convergence theorem but I can't find a larger series that is independent from x.

$\endgroup$
  • $\begingroup$ I guess, the $a_n$ and $b_n$ should be defined without the sum, right? $\endgroup$ – Michele Jul 12 '17 at 14:23
2
$\begingroup$

Using the inequality $\frac{2|ab|}{a^2 + b^2}\leq 1$ for every $a,b\in\mathbb{R}$ such that $(a,b)\neq (0,0)$, you get $$ \left| \frac{nx}{1+n^5 x^2} \right| = \frac{1}{2 n^{3/2}} \cdot \frac{2 n^{5/2} |x|}{1+ n^5 x^2} \leq \frac{1}{2 n^{3/2}}, \qquad \forall x\in\mathbb{R}, $$ hence the uniform convergence follows from the M-test.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ thank you for answering. your answer is great but I would like to know if what I did is wrong. what do you think? $\endgroup$ – segevp Jul 12 '17 at 14:50
  • 1
    $\begingroup$ The series $\sum 1/(n^4 x)$ is defined (and convergent) for every $x\neq 0$. Unfortunately, it is not uniformly convergent in $\mathbb{R}\setminus\{0\}$. $\endgroup$ – Rigel Jul 12 '17 at 14:53
  • $\begingroup$ I found out there is another way(the way they wanted me to solve it) if you derive it you can find a maximum point and you will get lim sup =0. $\endgroup$ – segevp Jul 12 '17 at 16:12
  • 1
    $\begingroup$ Yes, basically it's the same. In the first inequality, the equality sign (i.e. the max) is achieved when $|a|=|b|$, i.e. for $n^{5/2} |x| = 1$. Anyway, you obtain the same estimate. $\endgroup$ – Rigel Jul 12 '17 at 16:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.