functional analysis - Show that $f_varepsilonrightarrow 0$ as $varepsilonrightarrow 0$ in the space of basic functions

I have problem with this two problems:

1.
(already solved) What is an example of a sequence $\{x_n\}\in \ell_2$ (i.e. $\sum\limits_{n=1}^\infty x_n^2<\infty$) such that $\sum\limits_{n=1}^\infty \dfrac{x_n}{\sqrt{n}}=\infty$.
I have not a clue in this, since the series of the form $1/x^p$ does not work

2.
Let $K$ be the space of basic functions (functions $\phi:\mathbb{R}\rightarrow\mathbb{R},\phi\in C^\infty$ and such that exists $a\geq b$ such that $\phi=0\ \forall x\notin[a,b]$).

Let $f_\varepsilon=\dfrac{\varepsilon}{x^2+\varepsilon^2}$ be a sequence of functionals over $K$.

Show that $f_\varepsilon\rightarrow 0$ as $\varepsilon\rightarrow 0$. i.e. $(f_n,\phi)\rightarrow(0,\phi)\ \forall\phi\in K$

Note: $(f,\phi)=\int\limits_{-\infty}^\infty f(x)\phi(x)dx$

In this problem I have to whot that that
$$\varepsilon\int\limits_{-\infty}^\infty\dfrac{\phi(x)}{x^2+\varepsilon^2}dx\rightarrow0\text{ as }\varepsilon\rightarrow 0$$

Here $\int\limits_{-\infty}^\infty\dfrac{\phi(x)}{x^2+\varepsilon^2}dx=\int\limits_{a}^b\dfrac{\phi(x)}{x^2+\varepsilon^2}dx$ and since $\phi$ is continuous then $\dfrac{\phi(x)}{x^2+\varepsilon^2}$ is bounded for all $\varepsilon$, but is it uniformly bounded?

Answer

Hints:

1. The series $\sum_{n=2}^\infty {1\over n\log^pn}$, which converges if and only if $p>1$, might be useful.




2. Observe that ${d\over dx}\arctan(x/\epsilon)={\epsilon\over x^2+\epsilon^2}$, and integrate by parts.