Uniform Convergence of Series on [0,1]

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SUMMARY

The discussion focuses on proving the uniform convergence of the series \(\sum_{k=0}^{\infty} \frac{1-x}{(2+x)^k}\) on the interval [0,1]. Adam initially demonstrates that the limit as \(k\) approaches infinity of \(\frac{1-x}{(2+x)^k}\) equals 0, indicating pointwise convergence. To establish uniform convergence, it is essential to first confirm pointwise convergence and then apply the Weierstrass M-test, which utilizes the inequality \(|f_k(x)| \leq M_k\) to show that the series converges uniformly on the compact set [0,1].

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  • Study the Weierstrass M-test in detail to understand its application in proving uniform convergence.
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Pearce_09
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I wish I had more done.. but I'm have trouble understanding this...

Show that the series The sum is k=0 to infinity [tex]\sum^\infty_0 1-x/(2+x)^k[/tex] converges uniformly on [0,1].

What I have done so far is take the
the lim is k-->infinity [tex]\lim_\infty 1-x/(2+x)^k = 0[/tex]
therefore the limit exists...
does this even matter.. i mean, do I even need to state this.. because I still need to say if it converges on [0,1] which I am have trouble starting on.. please help.
thank you,
sorry i havnt quite mastered latex
adam
 
Last edited:
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Sloppiness with your parentheses!
(I take it you meant [itex](1-x)/(2+x)^{k}=\frac{1-x}{(2+x)^{k}}[/itex])

Hmm.. what you have shown so for is that for any fixed x, the series does not necessarily diverge.

As for proving uniform convergence, split it in two:

1. Show that you have pointwise convergence.
2. Having 1, since you have pointwise convergence on a compact set, then you also have uniform convergence on that set.
 
Last edited:
this is just a test
suppose [tex]|f_k(x)| \leq M_k[/tex]
then, [tex]\sum^\infty_1 |f_k(x)| \leq \sum^\infty_1 M_k[/tex]
 

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