# Convergence on [-M,M] for any M implies convergence on R?

Good day dear fellows. I am given the following series

$h(x) = \sum_{n=1}^{\infty} \frac{1}{x^2+n^2}.$

It is requested that I show that $h(x)$ is continuous on R. I did the following: use the Weirerstrass M-test to show uniform convergence, and then, using the continuity of the functions that are being summed, conclude that $h(x)$ is continuous. The solutions manual agrees with me. So far so good.

Next I am asked whether $h(x)$ is differentiable, and, if $h(x)$ is differentiable, to investigate if the derivative $h'(x)$ is continuous. I note that that

$\frac{d}{dx} \frac{1}{x^2+n^2} = \frac{-2x}{(x^2+n^2)^2}.$

At this point I don't see how I am going to get uniform convergence out of a summation over this thing, so I take out the solutions manual again. The manual uses the Weirerstrass M-test to show uniform convergence on $[-M,M]$ for arbitrary $M$, and concludes that there must be uniform convergence on R (because M is aribitrary). To be explicit: on any interval $[-M,M]$,

$|\frac{2x}{x^2+n^2} | \leq \frac{2M}{n^2}$.

(the square over the denominator can be removed since the denominator is greater than one).

And, since

$\sum_{n=1}^{infty} \frac{2M}{n^2}$

converges on [-M,M], uniform convergence of the derivatives follows via the weirerstrass M-test. Since $M$ is arbitrary, we then have uniform convergence (and thus continuity) on R.

I don't see how this is so, since R is not a closed interval..

Related Calculus and Beyond Homework Help News on Phys.org
HallsofIvy
I think I am confused by the notation for intervals. For any finite numbers $a$ and $b$, $(a,b)$ is an open interval, and every closed interval can be written $[c,d],$ where $c$ and $d$ are its endpoints. But this isn't true for unbounded intervals? In other words, $(-\infty, \infty)$ is a closed interval, since it contains all it's limit points (but not its endpoints)? Endpoints of unbounded intervals are not limit points?