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A function $f$ is defined as follow:

$$ f(x)=\sum_{n=1}^{\infty}\frac{b_{n}}{(x-a_{n})^{2}+b_{n}^{2}}\;\;, x\in \mathbb R $$

where $(a_{n}, b_{n})$ are points in the $xy$-plane, $b_{n}>0$ for all $n$. When is the function $f(x)$ bounded away from zero? that is $f(x)\geq a>0$, for some $a>0$, for all $x\in \mathbb R$. I believe that this would somehow depend on the points $(a_{n}, b_{n})$, for example if $\{b_{n}\}$ converges to zero or not, but I cannot find out how!


*EDIT: I still don't know when $f(x)$ will be bownded away from zero! *

Edit: $f(x)$ is bounded above by some constant, say $C>0$.

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Instead of "strictly positive" you can say: "bounded away from zero". – GEdgar Feb 20 '12 at 0:51
Yes, thanks! I fixed it. – Chelsea Feb 20 '12 at 1:10

All summands are strictly positive, so if the series converges (or if you accept $+\infty$ as strictly positive) the sum is strictly positive.

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So this means that the case $\lim_{x\rightarrow x_{0}} f(x)=0$ is not possible for any $x_{0}\in \mathbb R$?! – Chelsea Feb 20 '12 at 0:42
Of course; again, for any summand the limit is strictly positive. – Robert Israel Feb 20 '12 at 0:53
Changing "strictly positive" to "bounded away from zero" changes the question completely. – Robert Israel Feb 20 '12 at 4:41
My fault! I meant bounded away from zero. – Chelsea Feb 20 '12 at 17:23

Here's one in the opposite direction. Note that for any $c > 0$ the curve $y/(x^2 + y^2) = c$ is a circle in the upper half plane, tangent to the $x$ axis at the origin. If there is some $r > 0$ such that every circle of radius $r$ with centre on the line $y=r$ contains at least one $(a_n, b_n)$, then $f(x)$ is bounded below.

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How about $$ \lim_{x\to-\infty}\sum_{n = 0}^{\infty} \frac{1}{(x-n)^2 + 1} = 0 $$

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And somewhat more generally, if there is $x_0$ such that all $a_n \ge x_0$ and $\sum_{n=1}^\infty \frac{b_n}{(x_0 - a_n)^2 + b_n^2} < \infty$, then $\lim_{x \to -\infty} \sum_{n=1}^\infty \frac{b_n}{(x-a_n)^2 + b_n^2} = 0$ so in that case your sum is not bounded away from $0$. – Robert Israel Feb 20 '12 at 5:42
@Robert Israel: We can assume that $f(x)$ is bounded above, but does the $b_{n}$'s play any role here? – Chelsea Feb 20 '12 at 13:27
How do you see that the limit is zero? – Chelsea Feb 21 '12 at 19:57
@Chelsea: take $x<-k$, $k\in\mathbb{N}$. Then $\sum_{n = 0}^{\infty} \frac{1}{(x-n)^2 + 1}\leq\sum_{n = k}^{\infty} \frac{1}{n^2 + 1}$. – Martin Argerami Feb 25 '12 at 16:29
@Chelsea: Note that if $x<x_0$, then the smaller $x$ the smaller $f(x)$ (because every summand gets smaller). If you want it $f(x)$ to be smaller than $\varepsilon$, then you find $N$ such that $\sum_{n=N}^\infty \frac{b_n}{(x_0-a_n)^2 + b_n^2} < \varepsilon/2$ and look for $x$ so close to $-\infty$, that the beginning of the series gets smaller than $\varepsilon/2$. – savick01 Feb 25 '12 at 16:43

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