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## Series and continuous functions

Real Analysis
Papapetros Vaggelis
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### 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.
Riemann
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### 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:
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.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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### 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
Riemann
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### Re: Series and continuous functions

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

The condition 2 is not true.
The sequence must be decreasing
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$