Uniform Convergence of the series : $\sum\frac{1}{1+x^n}, x\ge0$

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It is clear that the series converges uniformly on the interval $[a,\infty)$ , for $a>1$ as follows on using the Weirstrass M-Test as

$$\frac{1}{1+x^n} < \frac{1}{a^n}$$

Kindly suggest what about rest of the domain. My book says discuss the convergence and uniform convergence. I am confused how to discuss the two concepts in this example. Please help.

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Let's denote $u_n(x)=\frac{1}{1+x^n}$ then it's clear that if $0\leq x\leq 1$ then $u_n(x)\not \to_{n\to\infty}0$ then $\sum_n u_n(x)$ is divergent and if $x>1$ then $u_n(x)\sim x^{-n}$ so $\sum_n u_n(x)$ is convergent by comparaison with the geometric series and hence the series $\sum_n u_n(x)$ is pointwise convergent on the interval $(1,+\infty)$.

Now if $x\geq a$ for $a>1$ we have $$||u_n||_\infty\leq a^{-n}$$ hence the series $\sum_n ||u_n||_\infty$ is convergent then the series $\sum_n u_n(x)$ is normal convergent and then uniformly convergent on the interval $[a,+\infty)$.

Recall We have: the normal convergence implies the uniform convergence implies the pointwise convergence but the converse isn't true.