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

a) Prove that $f$ is defined and continuous on $(-\infty,\infty)$ (don't use part (b) here)

Since for $x\in\mathbb{R}$ and $n=1,2,3,\ldots$ $$\frac{1}{x^2+n^2}\le \frac{1}{n^2}$$ and $\sum_1^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ we can conclude that the series is uniformly convergent (Weierstrass test). Since each $f_n(x)=\frac{1}{x^2+n^2}$ is continuous, the limit $f(x)$ is also continuous.

b) Prove that $f$ is differentiable on $(-\infty,\infty)$.

Let's consider interval $[-a,a]$ where $a$ is any positive real number. Each $f_n$ is differentiable on $\mathbb{R}$ and its derivative is $$f_n'(x)=\frac{-2x}{(x^2+n^2)^2}$$ For $x\in[-a,a]$ the following inequalities are true: $$ \Bigg| \frac{2x}{(x^2+n^2)^2}\Bigg|\le \frac{|2x|}{n^4}\le \frac{2a}{n^2}$$ We conclude that $\sum_1^{\infty} f'_n(x)$ converges uniformly on each $[-a,a]$, hence on all $\mathbb{R}$. It is also clear that $\sum_1^{\infty}f_n(0)$ is convergent. Having the above two condition satisfied, we know that $f$ is differentiable. Moreover, $f'(x)=\sum_1^{\infty} f'_n(x)$

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Your argument is correct until this point "We conclude that $\sum_1^{\infty} f'_n(x)$ converges uniformly on each $[-a,a]$, hence on all $\mathbb{R}$.". This is NOT correct. Uniform convergence on every closed bounded subset does not imply uniform convergence on $\mathbb{R}$. For example $f_n=\sum_{k=0}^n \frac{x^k}{k!}$ is uniform convergent on $[-a,a]$ for all $a>0$ but not on $\mathbb{R}$ ($f_{n}-f_{n-1}=\frac{x^n}{n!}$ is unbounded on $\mathbb{R}$.)

But to prove that statement, the uniform convergence on every closed bounded subset is enough. You just argue the differentiablity of $f$ on $[-a,a]$ for any $a>0$, hence $f$ is differentiable on $\mathbb{R}$.

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  • $\begingroup$ Good point, thanks! $\endgroup$ – luka5z Feb 4 '17 at 16:21

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