3
$\begingroup$

1 $$\sum_{n=1}^\infty (-1)^n \frac{x^2 + n}{n^2} $$ Is the series converges uniformly $\mathbb R$

I have tired by this result

  • if $\{f_n(x)\}$ is a sequence of a function defined on a domain $D$ such that

    1. $f_n(x) \geq 0$ for all $x \in D$ and for all $n \in \mathbb N$

    2. $f_{n+1}(x) \leq f_n(x)$ for all $x \in D$

    3. $\sup_{x\in D} \{f_n(x)\} \to 0$ as $n \to \infty$. Then

$\sum_{n=1}^\infty (-1)^{(n+1)}f_n(x)$ converges uniformly on $D$

first two condition this series is satisfied but third is not satisfied, so this is not work

  1. $$ \sum_{n=1}^\infty \frac{x \sin \sqrt{\frac{x}n}}{x +n} $$ Is the series converges uniformly on $[1, +\infty)$

I have tried Abel test and Dirichlet test,but not getting any solution.

  1. Study the uniform convergence of the series on $\mathbb R$ $$ \sum_{n=1}^\infty \frac{x\sin(n^2x)}{n^2} $$

Suppose this series converges uniformlybto $f$, so for a given $\epsilon > 0$, there exist $N \in \mathbb N$ such that $\left|\sum_{n=1}^k f_n(x) - f(x)\right| < \epsilon$ for all $k \geq N$

Please tell me how to proceed further. Any help would be appreciated , Thank you

$\endgroup$
  • $\begingroup$ I would suggest that you split your question in 3 questions in order to have separated answers. $\endgroup$ – mathcounterexamples.net Jan 13 '16 at 17:58
  • $\begingroup$ @ mathcounterexamples: I want to know that what is the basic tecnique use to get the answer in the unbounded domain, so i had asked inn one question and thanks for giiving suggestion. $\endgroup$ – Struggler Jan 14 '16 at 0:35
  • $\begingroup$ One of the basic techniques is to let $g_m(x)=\sum_1^nf_n(x)$ and consider that we have uniform convergence iff $\lim_{m\to \infty}\sup_{n>m}|\sup_x |g_n(x)-g_m(x)|=0$. In particular it is NECESSARY (but not sufficient) that $ \sup_x|f_{m+1}(x)|,$ which is $\sup_x|g_{m+1}(x)-g(x)|,$ must go to $ 0$ as $m\to \infty.$ This condition is not met in your Q so convergence is not uniform. $\endgroup$ – DanielWainfleet Jan 14 '16 at 1:27
3
$\begingroup$
  1. In general, if $\sum f_n(x)$ converges uniformly on $\mathbb R,$ then

$$ \sup_{\mathbb R}|f_n| \to 0 \text { as } n\to \infty.$$

This fails in your problem, because in this case $\sup |f_n| \ge |f_n(n^2)| = 1+1/n \not \to 0.$

  1. We can use the same idea as in 1. In this problem, $f_n(n) = (1/2)\sin 1 \not \to 0,$ so the series doesn't converge uniformly on $[1,\infty).$

  2. The series converges uniformly on any $[-a,a].$ Proof: We use Weierstrass M:

$$\sup_{[-a,a]}|f_n| \le \frac{a\cdot 1}{n^2}.$$

Since $\sum a/n^2 < \infty$ we have $\sum f_n$ converging uniformly on $[-a,a].$

However, the series does not converge uniformly on $\mathbb R.$ We can again use the idea in 1. Let $x_n = \pi/2n^2 + 2\pi n^2.$ Then verify

$$\sup_{\mathbb R} |f_n| \ge |f_n(x_n)|\ge 1 \not \to 0.$$

$\endgroup$
  • $\begingroup$ This is the more mature approach! Well done. +1 $\endgroup$ – Mark Viola Jan 13 '16 at 21:52
  • 1
    $\begingroup$ For $3$, with $x_n =\pi/ n^2+2\pi n^2$, $n^2x_n=\pi+2\pi n^4$ and $\sin (n^2x_n)=0$, does it not? Pehaps, $x_n=\pi/2n^2+2\pi n^2$. - Mark $\endgroup$ – Mark Viola Jan 13 '16 at 22:51
  • $\begingroup$ Yes you're right and that was what I intended. Thanks. $\endgroup$ – zhw. Jan 13 '16 at 23:25
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Jan 13 '16 at 23:30
  • $\begingroup$ @ Zhw :nice answer $\endgroup$ – Struggler Jan 14 '16 at 0:37
2
$\begingroup$

Regarding the first series $$\sum_{n=1}^\infty (-1)^n \frac{x^2 + n}{n^2} $$ the remainder $$R_{2n}(x) = \sum_{k=2n}^\infty (-1)^n \frac{x^2 + k}{k^2}$$ can be evaluated grouping one even term with one odd term. You get $$R_{2n}(x) = x^2\sum_{k=2n}^\infty \frac{2k+1}{k^2(k+1)^2} + \sum_{k=2n}^\infty \frac{1}{k(k+1)}$$ The second term of the RHS is converging to $0$ as the series $\sum \frac {1}{k(k+1)}$ is convergent.

Regarding the first one, we have $\frac{2k+1}{k^2(k+1)^2} \sim \frac{2}{k^3}$. Hence $$\sum_{k=2n}^\infty \frac{2k+1}{k^2(k+1)^2} \sim \frac{A}{n^2}$$ with $A > 0$. Therefore $R_{2n}(x)$ doesn't converge uniformly to zero (consider $x=n$). And the series is not uniformly convergent.

$\endgroup$
  • $\begingroup$ Well done, but there is a typographical error wherein $\sim \frac{A}{k^2}$ should read $\sim \frac{A}{n^2}$. ;-) ... - Mark $\endgroup$ – Mark Viola Jan 13 '16 at 18:24
  • $\begingroup$ Solid result. +1 ... - Mark $\endgroup$ – Mark Viola Jan 13 '16 at 18:37
2
$\begingroup$

For the second problem, we can use the inequality

$$\sin\left(\sqrt{\frac{x}{n}}\right)\ge \sqrt{\frac{x}{x+n}}$$

for $0\le \sqrt{x/n}\le \pi/2$.

Then, we have

$$\begin{align} \sum_{n=N}^\infty \frac{x\,\sin\left(\sqrt{\frac{x}{n}}\right)}{x+n}&\ge \sum_{n=N}^\infty \left(\frac{x}{x+n}\right)^{3/2}\\\\ &\ge \int_N^\infty \left(\frac{x}{x+y}\right)^{3/2}\,dy\\\\ &=\frac{2x^{3/2}}{\sqrt{x+N}}\\\\ &>1 \end{align}$$

whenever $x=N$ for $N\ge1$.

Therefore, there exists a number $\epsilon>0$ (here $\epsilon=1$ is suitable) so that for all $N'$ there exists an $x\in[1,\infty)$ (here $x=N$), and there exists a number $N>N'$ (here, take any $N>N'\ge 1$) such that $\left|\sum_{n=N}^\infty \frac{x\,\sin\left(\sqrt{\frac{x}{n}}\right)}{x+n}\right|\ge \epsilon$.

This is the statement of negation of uniform convergence and therefore the series fails to converge uniformly.

$\endgroup$
  • $\begingroup$ Thank you so much for giving your valuable time. $\endgroup$ – Struggler Jan 14 '16 at 0:39
  • $\begingroup$ You're welcome! My pleasure. $\endgroup$ – Mark Viola Jan 14 '16 at 0:39

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.