1
$\begingroup$

Which of the following are true

  1. $\sum_{n=1}^{\infty}\frac{(-1)^n + \frac{1}{2}}{n} $ is convergent.

I have tried by Lebnitz rule., but $|a_n|$ is not a decreasing sequence, So i can not use this, i have also tried to prove the series absolutely convergent, but not getting any sucees.

2.The series $$\sum_{n=1}^{\infty} \frac{x^2}{1+ n^2x^2}$$

converges uniformy on $\mathbb R$

I have tried by Weierstrass M-test, but $f_n(x) = \frac{x^2}{1 + n^2x^2}$ is unbounded on $\mathbb R$.Is i conclude that if $f_n(x)$ is unbounded , then $\sum f_n(x)$ is not uniformly convergent., but it is pointwise convergent.

Please tell me how to proceed.

  1. The series $$\sum_{n=1}^{\infty}\frac{\sin nx^2}{1+n^3}$$

converges uniformly on $\mathbb R$

Here $ |f_n(x)| \leq M_n = \frac{1}{1 + n^3}$ and $\sum M_n$ is convergent , then $\sum_{n=1}^{\infty} f_n(x)$ is unformly convergent by Weierstrass M-test.

Please tell me about (1) and (2) and check the (3) option.

Any help would be appreciated. Than you

$\endgroup$
4
  • $\begingroup$ 1) $\sum_{k=1}^N\frac{(-1)^n + \frac{1}{2}}{n} = \sum_{n=1}^N \frac{(-1)^n}{n} + \sum_{n=1}^N \frac{1}{2n}$. Do you know how to handle these two simpler series? $\endgroup$ – Winther Jan 10 '16 at 9:12
  • $\begingroup$ 2) Why do you say the function you give is unbounded? $\endgroup$ – πr8 Jan 10 '16 at 9:14
  • $\begingroup$ @ $\pi r 8$ : Since $f'_n(x) = \frac{2x}{(1 + n^2 x^2)^2}=0 $ , we got extreme points $x=0$, then $f_n(x)$ is minimum at $x=0$ but $f_n(x)$ has not maximum at any point. $\endgroup$ – user120386 Jan 10 '16 at 10:02
  • $\begingroup$ @winther: oh yes , i am confuse the point of rearrangements of the parenthesis of the series, but here no need. $\endgroup$ – user120386 Jan 10 '16 at 10:05
2
$\begingroup$

$\frac{x^2}{1+n^2x^2}<\frac{1/n^2+x^2}{1+n^2x^2}=1/n^2$ $\frac{(-1)^n+\frac{1}{2}}{n}=\frac{(-1)^n}{n}+\frac{1}{2n}$

$\frac{(-1)^n}{n}$‘s series is convergent,while $\frac{1}{2n}$’s series is +∞

$\endgroup$
2
  • $\begingroup$ @ Kai:Is this is a result if $f_n(x) \leq a_n$ and $\sum a_n$ is not convergent, then $\sum f_n(x)$ is not uniformly convergent. $\endgroup$ – user120386 Jan 10 '16 at 10:13
  • $\begingroup$ @user120386 obviouly no!......................................my answer means the series of the second problem of yours is convergent,while the first is not $\endgroup$ – Kai Jan 10 '16 at 11:08

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.