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Hi everyone: Suppose we have for two power series $$0\leq\sum_{n=0}^{\infty}a_{n}(x-\alpha)^{n}\leq \sum_{n=0}^{\infty}b_{n}(x-\alpha)^{n}$$ for all $|x-x_{0}|0$. The left hand series converges for $|x-x_{0}|r$. Can we conclude that the left hand series also converges for $|x-x_{0}|

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    Should it be $(x-\alpha)^n$ perhaps? Right now there is no power series involved here. – 2017-02-22
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    Yes, thank for letting me know. I corrected. – 2017-02-23

3 Answers 3

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$\frac 1{1+x^2} = \sum (-1)^{n} x^{2n}$ conveges when $r<1$ and diverges outside this radius.

$\sum (-1)^{n} x^{2n} < 1$ for all $x$ in $(-1,1)$

$\sum \frac {e x^n}{n!} = e^{x+1}$ converges for all $x$

$\sum \frac {e x^n}{n!} = e^{x+1} > 1$ for all $x\in (-1,1)$

No, your premise is not true.

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If the radius of converge of the $a_n$-series were smaller, say $r'

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Answer is no.

Here is a simple example.

Let $P$ be a polynomial that is positive on an interval $[-1,1]$. Clearly $P$ is a power series with radius of convergence equal to $\infty$.

Let $f(x) = \sum_{n=1}^\infty x^{2n}/n^2$. Note that $f$ is nonnegative and bounded on $[-1,1]$, but the power series diverges outside this interval (general term tends to $\infty$).

Then there exists a positive constant $c$, depending on $P$ and $f$ such that

$$0\le P-cf \le P\mbox{ on }[-1,1].$$

Now $P-cf$ is also a power series. Yet, its radius of convergence is $1$. It does not converge outside $[-1,1]$.