# Uniform convergence of $\sum\limits_{n=1}^{\infty}{\frac{x^{n}}{100+x^{2n}}}$

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.

• For $|x|>1$, this series diverges trivially, since its general term tends to $1$. – Bernard Mar 26 at 21:19
• 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? – TonyK Mar 26 at 21:47
• @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. – paul lacher Mar 26 at 21:56
• The image has $x^n$ in the numerator, not $x^{2n}$. – TonyK Mar 26 at 22:00
• @TonyK yes I realised, that was an error on my part – 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.
• For really big $n$, the constant term $100$ doesnt matter, i.e. for very large $n$ we have that $x^{2n} \approx x^{2n} +100$ – rubikscube09 Mar 26 at 21:46
• @paullacher: Just as a sanity check: what does $\frac{n}{n+1}$ tend to as $n\to\infty$? – TonyK Mar 26 at 21:50
• @paullacher $\displaystyle\lim_{n\to\infty}\frac{x^{2n}}{100+x^{2n}}=\lim_{n\to\infty}\frac1{100x^{-2n}+1}=\frac11=1.$ – José Carlos Santos Mar 26 at 21:53