Given the series:$\Sigma_{k=1}^{\infty} \frac{x^2 \cos (2x)}{1+k^4x^6} $ show that the series converge uniformly.
I tried to do it like this: $$\frac{x^2 \cos (2x)}{1+k^4x^6} \le \frac{x^2 }{1+k^4x^6} \\ \text{Atvasina pēc } x \text{ un pielīdzina nullei: } \frac{2x(1+k^4x^6) -x^2 (6k^4x^6) }{(1+k^4x^6)^2} = 0 \\ x(2+2k^4x^6-6k^4x^7) = 0 \Rightarrow x_1 = 0 \\ x(k^4x^5-3k^4x^6)=-1 \Rightarrow x_2 = -1 \\ x^5(k^4-3k^4x) = 0 \Rightarrow x_3 = -\frac{1}{3} \\ f''_k =\frac{(2+2k^4x^6+12x^6-26k^4x^7-36k^4x^7)(1+k^4x^6)-(2x(1+k^4x^6) -x^2 (6k^4x^6)) 6k^4x^5}{1+k^4x^6} \\ f''_k(0) = 2 > 0 \Rightarrow \text{ nav max} \\ f''_k(-1) = (2+4k^4+24 +52k^4 +72k^4) (1+2k^4) -(12k^4+24k^8-2\cdot 1296k^8) > 0 \\$$
But the way I do it, it looks clumsy and no point is max. Then what could be another way?
The $x^2\cos{(2x)}$ can be pulled out of the summation since the index of the summation is $k$. You're left with $\sum_{k=1}^{\infty} \frac{1}{1+k^4x^6}$, where x is treated as a constant. Compare the summation to $\sum_{k=1}^{\infty} \frac{1}{k^4x^6}$, which is larger than the original summation. The summation $\sum_{k=1}^{\infty} \frac{1}{k^4x^6}$ converges from the integral test or p-series test for $x\neq0$. By direct comparison test, the original summation converges. And for the case when $x=0$ the the summation turns into $\sum_{k=1}^{\infty} 0$ which converges.