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[itex] h(x) = \sum_{n=1}^{\infty} \frac{1}{x^2+n^2}. [/itex]

It is requested that I show that [itex]h(x)[/itex] 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 [itex] h(x) [/itex] is continuous. The solutions manual agrees with me. So far so good.

Next I am asked whether [itex] h(x) [/itex] is differentiable, and,

*if*[itex] h(x) [/itex] is differentiable, to investigate if the derivative [itex] h'(x) [/itex] is continuous. I note that that

[itex] \frac{d}{dx} \frac{1}{x^2+n^2} = \frac{-2x}{(x^2+n^2)^2}.[/itex]

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 [itex] [-M,M] [/itex] for arbitrary [itex]M[/itex], and concludes that there must be uniform convergence on R (because M is aribitrary). To be explicit: on any interval [itex] [-M,M] [/itex],

[itex] |\frac{2x}{x^2+n^2} | \leq \frac{2M}{n^2} [/itex].

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

And, since

[itex] \sum_{n=1}^{infty} \frac{2M}{n^2} [/itex]

converges on [-M,M], uniform convergence of the derivatives follows via the weirerstrass M-test. Since [itex]M[/itex] 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..