Welcome to mathimatikoi.org forum; Enjoy your visit here.
Series and continuous functions

 Community Team
 Articles: 0
 Posts: 426
 Joined: Mon Nov 09, 2015 1:52 pm
Series and continuous functions
Prove that the series \(\displaystyle{\sum_{k=2}^{\infty}\dfrac{\sin\,(k\,x)}{\ln\,k}}\) and the
series \(\displaystyle{\sum_{k=2}^{\infty}\dfrac{\sin\,(k\,x)}{k\,\ln\,k}}\) converge for each
\(\displaystyle{x\in\left[0,2\,\pi\right]}\).
Examine if the functions
\(\displaystyle{x\mapsto \sum_{k=2}^{\infty}\dfrac{\sin\,(k\,x)}{\ln\,k}\,,x\in\left[0,2\,\pi\right]}\)
and
\(\displaystyle{x\mapsto \sum_{k=2}^{\infty}\dfrac{\sin\,(k\,x)}{k\,\ln\,k}\,,x\in\left[0,2\,\pi\right]}\)
are continuous.
series \(\displaystyle{\sum_{k=2}^{\infty}\dfrac{\sin\,(k\,x)}{k\,\ln\,k}}\) converge for each
\(\displaystyle{x\in\left[0,2\,\pi\right]}\).
Examine if the functions
\(\displaystyle{x\mapsto \sum_{k=2}^{\infty}\dfrac{\sin\,(k\,x)}{\ln\,k}\,,x\in\left[0,2\,\pi\right]}\)
and
\(\displaystyle{x\mapsto \sum_{k=2}^{\infty}\dfrac{\sin\,(k\,x)}{k\,\ln\,k}\,,x\in\left[0,2\,\pi\right]}\)
are continuous.
Re: Series and continuous functions
Greetings to all,
I think the exercise is quite easy and we will be basing the solution on the following facts:
both serieses converge uniformly throughout $\mathbb{R}$ since observation $(2)$ holds and because the functions are continuous its uniform limit is a continuous function. Both these facts are quite well known results.
That's all folks.
I think the exercise is quite easy and we will be basing the solution on the following facts:
 The uniform limit of continuous functions is continuous.
 The series $\sum a_n \sin nx$ converges uniformly throughout $\mathbb{R}$ iff $n a_n \rightarrow 0$.
both serieses converge uniformly throughout $\mathbb{R}$ since observation $(2)$ holds and because the functions are continuous its uniform limit is a continuous function. Both these facts are quite well known results.
That's all folks.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1p^{s}}$

 Articles: 0
 Posts: 16
 Joined: Fri Aug 12, 2016 4:33 pm
Re: Series and continuous functions
The first function is not continuous.(no Lebesgue integrable)
The condition 2 is not true.
The sequence must be decreasing
The condition 2 is not true.
The sequence must be decreasing
Re: Series and continuous functions
Oh sorry , my bad!!S.F.Papadopoulos wrote:The first function is not continuous.(no Lebesgue integrable)
The condition 2 is not true.
The sequence must be decreasing
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1p^{s}}$