Exercise 6.4.4

Let $h_n(x)=\frac{x^{2n}}{(1+x^{2n})}$ be the terms being summed. For $\left|x\right|\ge 1$, $h_n(x)$ does not approach 0 and therefore the series does not converge. For $\left|x\right|<1$, $\left|h_n(x)\right|\le x^{2n}$, which forms a geometric series in $x^2$, which converges, so $g(x)$ converges by the Order Limit Theorem.

Note that for any $0\le a<1$, $\left|h(x)\right|\le a^{2n}=M_n$ over $[-a,a]$, and thus by the Weierstrass M-test $g(x)$ uniformly converges over $[-a,a]$ and is thus continuous over this interval. This last statement is equivalent to saying $g(x)$ is continuous over $(-1,1)$, which is also the set where $g(x)$ is well defined.