Power series: convergence on right-open interval

Question

Given the power series $f(x) = \sum_{k\geq 1} a_k x^k$ where $a_k \geq 0$, is the following statement true?

Let $f(x) = \sum_{k\geq 1} a_k x^k$ with $a_k \geq 0$. If $f(x) < \infty$ for some fixed $x \geq 0 $, then there exists some $\epsilon > 0$ such that $f(x + \epsilon) < \infty$.

In other words, the above states that the supremum of $\{x \geq 0: f(x) < \infty\}$ is not attained.

Answer 1

Let $a_k=k^k $ and $ x=0$. Conclusion : the statement is not true.

Answer 2

Fred already answered with a simple counter-example with $x=0$.

There is also a counter-example on the statement with $x > 0$ instead of $x \geq 0$.

Consider $a_k = 2^{2k} / k^2$, so \begin{equation} f(x) = \sum_{k=1}^\infty \frac{2^{2k} x^k}{k^2}\,. \end{equation}

This power series converges for $x \in [-1/4, 1/4]$ and diverges for $x < -1/4$ and $x > 1/4$.

To see this, you can apply the Ratio test to find

\begin{equation} \lim_{k\rightarrow\infty} \frac{|a_{k+1} x^{k+1}|}{|a_k x^k|} = 4 |x|\,, \end{equation}

proving convergence in $x \in (-1/4, 1/4)$ and divergence for $x < -1/4$ and $x > 1/4$. One can then check the convergence of the end points separately:

\begin{equation} f(1/4) = \sum_{k=1}^\infty \frac{1}{k^2} < \infty\,, \end{equation} and \begin{equation} f(-1/4) = \sum_{k=1}^\infty \frac{(-1)^k}{k^2} < \infty\,. \end{equation} (The latter series converges because the absolute value series converges.)

Full details can be found in: http://sites.millersville.edu/bikenaga/calculus/intervals-of-convergence/intervals-of-convergence.html

Answer 3

Consider $a_k=1/k^2$ and $x=1$.