Please show that $f$ is differentiable over any interval of the type $(a,\infty)$ with $a>0$.
$$f(x)=\sum_{k=1}^{\infty} \frac{1}{1+k^6x^2}. $$
Attempt: I figured that since we are on Chapter 7 in Rudin now covering sequences of functions and series, I should use the Weierstrass M-test.
So I took the derivative of $f$, and found $f'(x)$ to be $$\sum_{k=1}^{\infty} \frac{-2k^6x}{(1+k^6x^2)^2} $$. Note I think the lower bound of the sum may start from 2 instead of 1 (not sure). Anyway, I try to bound this so that I can use the Weierstrass M-Test. I get: $$|\sum_{k=1}^{\infty} \frac{-2k^6x}{(1+k^6x^2)^2}| \leq \sum_{k=1}^{\infty} \frac{2k^6x}{(1+k^6a^2)^2}$$.
The problem is I don't know how to get a bound for the top. I have this x on the top and I want my bound or $M_n$ to not depend on x. This is the part , I get stuck on. Any help would be much appreciated. Thank you. I know that once I get my $M_n$ , I have to show my series of $M_n$'s converges.
Note that $$ \frac{2k^6x}{(1+k^6x^2)^2}=\frac{2k^6}{x^{-1}+2k^6x+k^{12}x^3} $$ and that $$ x^{-1}+2k^6x\ge 2\sqrt{2}k^3>0 $$ Therefore, $$\frac{2k^6x}{(1+k^6x^2)^2}\le\frac{2k^6}{k^{12}x^3}\le\frac{2}{a^3k^{6}}$$