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How do i prove the function:

$g(x)=\sum_{n=1}^{\infty }\frac{1}{^{n^{0.5}}}(x^{2n}-x^{2n+1})$

is continuous in [0,1]?

I tried to look at this functions as:

$g(x)=(1-x)\sum_{n=1}^{\infty }\frac{1}{^{n^{0.5}}}x^{2n}$

but I couldn't find a way solving it...

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Your TeX is not making sense, $\sqrt{^n}$ –  GEdgar Jan 25 '13 at 15:03
    
my mistake... now it's supposed to be fine –  user59640 Jan 25 '13 at 15:05
    
Did you try to calculate the radius of convergence? –  Fabian Jan 25 '13 at 15:13
    
Yeah it is 1... But it just proves that the function is continuous in every closed interval [0,r] while 0<r<1, no? –  user59640 Jan 25 '13 at 15:14
    
It is continuous on every closed interval in $(-1,1)$. So the only thing left to proof is the continuity at $1$. –  Fabian Jan 25 '13 at 17:58

1 Answer 1

In fact, the series converges uniformly on $[0,1]$. We can show $$ \frac{x^{(2 n)} - x^{(2 n + 1)}}{\sqrt{n}} \le \frac{\Bigl(\frac{2n}{2 n + 1}\Bigr)^{2 n}}{\sqrt{n} (2 n + 1)} \approx \frac{1}{2 e n^{3/2}} $$ and $\sum 1/(2en^{3/2})$ converges.

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