Let $a_n$ a sequence such that $\lim\limits_{k\to\infty}a_n= +\infty$ and $\sum 1/|a_n|$ converges. Prove that the series of functions given by:
$$f(x)=\sum_{n=1}^{\infty} 1/(x-a_n)$$
converges absolutely and uniformly in every closed interval $X$ such that $X$ does not contains any $a_n$.
My attempt: Let $N\in\mathbb{N}$ such that $a_N\not=0$, and $X$ and closed interval such that $X$ does not contains points of $a_n$ Since $a_n$ tends to $\infty$ exists a natural $N_1$ such that for every $n\geq{N_1}$ we have that $|a_n-x|\geq{|a_N|}$, and $x\in{X}$, therefore, $1/|a_n-x|\leq{1/|a_N|}$ for every $n\geq{N_1}$. By hypothesis $\sum_{n=N}^{\infty} 1/|a_n|$ converges therefore by Weierstrass theorem we can conclude that $f(x)=\sum_{n=1}^{\infty} 1/(x-a_n)$ converges absolutely and uniformly in $X$.
Is my proof correct? Thanks for any help.
Best Answer
You are comparing $|a_n-x|^{-1}$ to $|a_N|^{-1}$ but this latter is a constant number, you cannot sum it, this will diverge.
Yet the idea is fine, you can rewrite $\dfrac 1{|x-a_n|}=\dfrac 1{|a_n||1-\frac x{a_n}|}$
Now since $x$ is fixed and $a_n\to\infty$ there is some $N\in\mathbb N$ such that $|1-\frac x{a_n}|>\frac 12$ for $n>N$.
And you can conclude by comparison with $\sum \frac 1{2|a_n|}$ for absolute convergence.
But I don't think the convergence is uniform over $X$, if $X$ is not bounded the $N$ found depends on $x$.