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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!

Thanks

*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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    Yes, thanks! I fixed it.2012-02-20

3 Answers 3

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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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    My fault! I meant bounded away from zero.2012-02-20
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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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    Note also, that \sum \frac{1}{b_n} < \infty also quarantees that $\lim_{|x| \to \infty} \sum \frac{b_n}{(x-a_n)^2 + b_n^2} = 0$. Suprisingly the proof is exactly the same as above.2012-02-25