# Uniform convergence

#### flower3

Discuss the Uniform convergence of the following series of functions :
1- $$\displaystyle \sum \frac{x^n}{1+ x^n} \ on \ [0,a] , a<1 \ and \ on \ [0,1)$$
2- $$\displaystyle \sum (-1)^n x^n (1-x) \ on \ [0,1]$$
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1- on [0,a] :
BY Weierstrass M- Test :
$$\displaystyle \left | \frac{x^n}{1+x^n} \right |\leq \frac{a^n}{1}$$
$$\displaystyle \sum a^n$$ is conv {geometric series }
$$\displaystyle \rightarrow$$ $$\displaystyle \sum \frac{x^n}{1+ x^n}$$ is Uniformly convergent on [0,a]
on [0,1):
?????
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2-

#### chisigma

MHF Hall of Honor
The series...

$$\displaystyle S = \sum_{n=0}^{\infty} (-1)^{n} x^{n} (1-x)$$ (1)

... converges is $$\displaystyle [0,1]$$. For $$\displaystyle x=1$$ all terms os (1) vanish and is $$\displaystyle S=0$$ and for $$\displaystyle 0 \le x <1$$ is...

$$\displaystyle S=\sum_{n=0}^{\infty} (-1)^{n} x^{n} (1-x)= \frac{1-x}{1+x}$$ (2)

Now if we compute the remainder...

$$\displaystyle R_{k} = \sum_{n=k}^{\infty} (-1)^{n} x^{n} (1-x) = (-1)^{k} x^{k} \frac{1-x}{1+x}$$ (3)

... for any $$\displaystyle \varepsilon >0$$ the inequality $$\displaystyle |R_{k}|< \varepsilon$$ is verified only if is ...

$$\displaystyle k \ln x < \ln \varepsilon + \ln (1+x) - \ln (1-x) \rightarrow k> \frac{\ln \varepsilon + \ln (1+x) - \ln (1-x)}{\ln x}$$ (4)

But the limit for $$\displaystyle x \rightarrow 1-0$$ of the (4) is unbounded so that the (1) doesn't converge uniformely in $$\displaystyle [0,1)$$ ...

Kind regards

$$\displaystyle \chi$$ $$\displaystyle \sigma$$

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