LINK FOR EXERCISEProve that $$ \sum_{n=1}^\infty f_n(x) = \sum_{n=1}^{\infty}{\frac{x^{ n}}{100+x^{2n}}}$$ converges uniformly for $x \in [2,\infty)$? I know it converges uniformly, but when I tried to prove it, I get stuck when trying to reformat it into a geometric series in order to find the limit, but I know it would be better to apply Weierstrass M-test but the $M_n$ that I found to be larger or equal to the sum above would be $$\sum_{n=1}^{\infty} {\frac{x^{n}}{100+x^{n}}}$$ but I am not sure how to prove that is converges, as by ratio and root test, it doesn't work out very well.

*edit: I made a typo, the start of summation is at n=1 not n=0, as well in my exercise sheet, it said to prove that it converges uniformly , sorry about that. *edit #2, I made another typo, for the numerator it is supposed to be $x^n$ not $x^{2n}$ sorry.

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    $\begingroup$ For $|x|>1$, this series diverges trivially, since its general term tends to $1$. $\endgroup$ – Bernard Mar 26 at 21:19
  • $\begingroup$ You say you know it converges uniformly; but this is manifestly false, as others have pointed out. So how do you "know" it? Perhaps you have copied it wrong? $\endgroup$ – TonyK Mar 26 at 21:47
  • $\begingroup$ @TonyK Maybe there was a mistake on the exercise sheet, but it said to prove that it converges uniformly, I will try to link an image in the post. $\endgroup$ – paul lacher Mar 26 at 21:56
  • $\begingroup$ The image has $x^n$ in the numerator, not $x^{2n}$. $\endgroup$ – TonyK Mar 26 at 22:00
  • $\begingroup$ @TonyK yes I realised, that was an error on my part $\endgroup$ – paul lacher Mar 26 at 22:02

If$$f_n(x)=\frac{x^n}{100+x^{2n}},$$then$$f_n'(x)=\frac{n x^{n-1} \left(100-x^{2n}\right)}{\left(x^{2 n}+100\right)^2},$$which is negative, for every $x\geqslant2$, if $n\geqslant4$ (because then $x^{2n}\geqslant2^8=256$). So, for $N\geqslant4$, $f_n$ is decreasing and therefore$$f_n(x)\leqslant f_n(2)=\frac{2^n}{100+4^n}<\frac1{2^n}.$$But the series $\sum_{n=4}^\infty\frac1{2^n}$ converges. Therefore, your series converges uniformly, by the Weierstrass $M$ test.

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  • $\begingroup$ thank you, but I am a bit confused, how would it go to 1, wouldn't it go to 0 as the denominator is larger than the numerator? $\endgroup$ – paul lacher Mar 26 at 21:35
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    $\begingroup$ For really big $n$, the constant term $100$ doesnt matter, i.e. for very large $n$ we have that $x^{2n} \approx x^{2n} +100$ $\endgroup$ – rubikscube09 Mar 26 at 21:46
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    $\begingroup$ @paullacher: Just as a sanity check: what does $\frac{n}{n+1}$ tend to as $n\to\infty$? $\endgroup$ – TonyK Mar 26 at 21:50
  • $\begingroup$ @TonyK it's 1 no? $\endgroup$ – paul lacher Mar 26 at 21:52
  • $\begingroup$ @paullacher $\displaystyle\lim_{n\to\infty}\frac{x^{2n}}{100+x^{2n}}=\lim_{n\to\infty}\frac1{100x^{-2n}+1}=\frac11=1.$ $\endgroup$ – José Carlos Santos Mar 26 at 21:53

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