Uniform Convergent and term by term differentiation

Question

Show that $f(x)= \sum_{n=1}^{\infty} \frac{1}{1+x^2+n^4x^2}$ is uniformly convergent for all real values of $x$.

Examine whether $f'(0)$ can be found by term-by-term differentiation.

Answer 1

$\sum_{n=1}^{\infty} \frac{1}{1+x^2(n^4+1)}$ converges pointwise for all $x \ne 0$.

The convergence is not uniform because for $x_m = \frac1{\sqrt{m^4+1}}$ we have

$$ \sup_{x\in\mathbb{R}}\left|\sum_{n=m}^{\infty} \frac{1}{1+x^2(n^4+1)} \right| \ge \frac{1}{1+x_m^2(m^4+1)} = \frac12$$

which doesn't converge to $0$ as $m\to\infty$.

Your function $f$ is not defined at $0$ so $f'(0)$ doesn't make sense. However, term-wise differentiation at $x = 0$ gives

$$-\sum_{n=1}^\infty \frac{2x(n^4+1)}{(1+x^2(n^4+1))^2}\Bigg|_{x = 0} = \sum_{n=1}^\infty 0 = 0$$