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

  • $\begingroup$ 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

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

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

  • $\begingroup$ @ 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

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