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.
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.