Is $\sum_{j=0}^\infty \frac{x}{(1+x^2)^j}$ pointwise convergent or uniform convergent?

Question

My initial thoughts for this question was to find the Chebyshev norm of $f_j(x) = \frac{x}{(1+x^2)^j}$. Doing this I get $\vert\vert f_j \vert\vert _\infty = 1$. Does this mean that the series is pointwise convergent to the constant function 1 and how would I go from here to determine whether it is uniformly convergent?

Answer 1

Note that $x$ is a constante here, with respect to $j$. Therefore\begin{align}\sum_{j=0}^\infty\frac x{(1+x^2)^j}&=x\sum_{j=0}^\infty\left(\frac1{1+x^2}\right)^j\\&=\frac x{1-\frac1{1+x^2}}\\&=\frac{x+x^3}{x^2}\\&=x+\frac1x,\end{align}unless $x=0$, in which case the sum is $0$, of course. Since the sum is discontinuous but each function is continuous, the convergence cannot be uniform.