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

  • $\begingroup$ 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
  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

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

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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