Show that $\sum_{n=0}^{\infty}{\frac{n^2 x}{1+n^4 x^2}}$
converges uniformly on $[\delta,1]$ where $0<\delta<1$.
I tried to use $M$ test.
$$\frac{n^2 x}{1+n^4 x^2} \leqslant \frac{n^2 x}{n^4 x^2} = \frac{1}{n^2 x}$$ Then, can I say that: $$\sum_{n = 1}^{\infty}\frac{1}{n^2x}$$ converges if I prove that: $$\sum_{n = 1}^{\infty}\frac{1}{n^2}$$ is convergent?? If so, it would prove the theorem right? So here, do I treat $x$ in this case as a constant??
I am confused...
Since $x \in [\delta ,1]$ $1/x \le 1/ \delta$. Hence $\sum_{n=1}^\infty\frac 1 {n^2x} \le \frac 1 \delta\sum_{n=1}^\infty \frac 1 {n^2}$ and you get your result by the M-test.