Exercise 7.4

Let $f_n(x)$ be the $n^{\mathit{th}}$ term of the series. Since $f_n(0)=1$ for all $n$, the series diverges for $x=0$. If $x>0$, then

by Theorem 3.28, so the series converges if $x>0$. The convergence is absolute since $f_n(x)>0$ for $x>0$, and since $f_n(x)\to \infty$ as $x\to 0+$, $f(x)$ is not bounded on $(0,\infty )$. If $a>0$, then since $f_n(x)\le 1∕(a{n}^2)$ for $a\le x$ and all $n$, the series converges uniformly to $f(x)$ on $[a,\infty )$ by Theorem 7.10. Since the partial sums are continuous, $f(x)$ is continuous on $[a,\infty )$ by Theorem 7.12, so it is continuous on all of $(0,\infty )$. The series does not converge uniformly on $(0,\infty )$ since the difference between $f(x)$ and a partial sum is a series ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_N^{\infty }{f}_n(x)>1∕(1+N^{2}x)$, which is arbitrarily large as $x\to 0+$.

The case $x<0$ is more complicated. First note that $f_n(x)$ is not even defined for $x=-n^{-2}$, so we are limited to considering the intervals $(-\infty ,-1)$ and $\left(-n^{-2},-{(n+1)}^{-2}\right)$ for $n=1,2,\dots$ For $-n^{-2}<x<-{(n+1)}^{-2}$ and $m>n+1$ we have

so that

Similarly, for $x<-1$ and $m>1$, we get

which is the same inequality, with $n=0$. The series

converges, which can be shown either by the integral test or the “limit comparison test,” which was not covered in Rudin’s text. Hence the series converges uniformly and absolutely for $-n^{-2}<x<{(n+1)}^{-2}$, or $x<-1$ by Theorem 7.10. Since the partial sums are continuous, $f(x)$ is continuous on this interval by Theorem 7.12. It is not bounded, since $f_n(x)\to -\infty$ as $x\to -n^{-2}$, while the rest of the series (everything except the $n^{\mathit{th}}$ term) sums to a finite value.