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How can i show uniform convergence on function series like this one: $\sum\limits_{k=1}^{\infty} (\sqrt{1-x^{n}}-1)$ ? I have a given interval of [0 / 0.5]

I thought about using the Weierstrass M-test, but i failed in finding the right sequence of numbers.

At the moment, i think, it is impossible to show convergence for the given function series.

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  • $\begingroup$ You need the $k$ and $n$ to coincide, yes? $\endgroup$ – Pedro Tamaroff May 4 '13 at 19:15
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You know by Taylor's theorem that $\sqrt{1-y}-1=-\frac{y}{2}+O(y^2)$, or in other words, there exists some $C,\delta>0$ such that $|\sqrt{1-y}-1|\leq \frac{|y|}{2}+Cy^2$ for all $|y|<\delta$. In your case, plug in $y=x^{n}$, so that for $0<x<0.5$, there is some $N>0$ (which depends only on the interval $x$ lives in: $[0,0.5]$) such that $|\sqrt{1-x^n}-1|<Cx^n$ for all $n\geq N$. Can you complete the rest?

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Have you tried to show normal convergence ? It's a stronger result, but, when it holds, it is also often easier to prove.

In your case, $$\sup_{x\in[0,1/2]}|\sqrt{1-x^n}-1|=1-\sqrt{1-\frac{1}{2^n}}\sim\frac{1}{2^{n+1}}$$

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  • $\begingroup$ Oh, i didn't try yet, but i will. for growing n the value will fall towards Zero. But, the series given in my example has negative values, that "rise" towards Zero with growing n. I am kinda confused. $\endgroup$ – Toralf Westström May 4 '13 at 20:32
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    $\begingroup$ Follow the link to have a definition of normal convergence. Basically, if the series $x\mapsto\sum f_n(x)$ is such that $\sum_{n=0}^\infty||f_n||_\infty<\infty$, then $\sum f_n(x)$ converges normally (and therefore also uniformly). Whether $f_n$ takes positive or negative values is irrelevant. $\endgroup$ – Clement C. May 4 '13 at 20:52

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