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Series and continuous functions

Real Analysis
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Papapetros Vaggelis
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Series and continuous functions

#1

Post by Papapetros Vaggelis » Fri Mar 24, 2017 3:14 pm

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.
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Riemann
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Re: Series and continuous functions

#2

Post by Riemann » Fri Mar 24, 2017 6:05 pm

Greetings to all,

I think the exercise is quite easy and we will be basing the solution on the following facts:
  1. The uniform limit of continuous functions is continuous.
  2. The series $\sum a_n \sin nx$ converges uniformly throughout $\mathbb{R}$ if-f $n a_n \rightarrow 0$.
Now,

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.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
S.F.Papadopoulos
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Re: Series and continuous functions

#3

Post by S.F.Papadopoulos » Thu Mar 30, 2017 7:53 pm

The first function is not continuous.(no Lebesgue integrable)

The condition 2 is not true.
The sequence must be decreasing
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Re: Series and continuous functions

#4

Post by Riemann » Mon Apr 03, 2017 10:29 am

S.F.Papadopoulos wrote:The first function is not continuous.(no Lebesgue integrable)

The condition 2 is not true.
The sequence must be decreasing
Oh sorry , my bad!! :oops: :oops:
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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