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Let $x$ be a small real number, say $1/10 <x< 3/10$. Let $(a_n)$ be a sequence of integers with $\vert a_n\vert \leq 3^n$. How can I find a positive lower bound for $$\left\vert\sum_{n=1}^\infty a_n x^n\right\vert$$ assuming that the function $\sum_{n=1}^\infty a_n z^n$ has no zeroes in $\mathbf{R}-\{0\}$.

Using the triangle inequality it's easy to find an upper bound for the above sum, but I don't know any method for finding a lower bound.

Would a lower bound like $\vert a_n\vert \geq 2^n$ help?

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I don't understand what restrictions you are looking for on $(a_n)$. You could use things like $\left|\sum\limits_{n=1}^\infty a_n x^n\right|\geq\left|\sum\limits_{n=1}^m a_n x^n\right| -\left|\sum\limits_{n=m+1}^\infty a_n x^n\right|$. – Jonas Meyer Dec 4 '11 at 19:06
Yes, this way I just have to bound the last term from above. This can be done explicitly using the triangle inequality. In fact, the last term you write down is bounded from above by $(3x)^m/(1-3x)$. But how do I bound the first term from below? My only guess is that I require the exact values of $a_n$ for $n=1,\ldots,m$ and then use a computer to approximate this number. – Federico Dec 4 '11 at 19:25
@Federico: With only an upper bound for the $a_n$ and the information that $f(x) = \sum_{n=1}^\infty a_nx^n$ has not zeroes except at $x=0$, you can always have something like $f(x) = a_1 x$ with arbitrarily small $|a_1|>0$. So you definitely need more information, i.e. a lower bound for the $a_n$ to be able to give a lower bound for $f$. In this case, Jonas Meyer already has poitned out the way to go. – Sam Dec 4 '11 at 19:42
@Sam. Your comment convinced me that I will need a lower bound on the $a_n$. But I still dont know how to find a lower bound for the term $\vert \sum_{n=1}^m a_n x^n\vert$ in Jonas' answer. I can only think of using a computer, but you can only compute a finite number of values of $f$ this way. Anyway, as I said in my above comment, I know that $\vert \sum_{n= m+1}^\infty a_n x^n\vert \leq (3x)^m/(1-3x)$, but how do I get a lower bound for $\vert\sum_{n=1}^\infty a_n x^n\vert$? – Federico Dec 4 '11 at 20:46
You could set $m=1$ and get $\left|\textstyle\sum_{n=1}^\infty a_nx^n \right| \ge \left|a_1x \right| - \left|\textstyle\sum_{n=2}^\infty a_nx^n \right|$ for example. But it always depends on what exactly you're trying to do... – Sam Dec 6 '11 at 0:09

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