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let a power series be:$$ \sum_{n=0}^{\infty} a_n\left(x-a\right)^n $$

Then there exists an R with 0<=R<=infinity such that

  1. the series converge if $\;|x-a| \le R\;$
  2. the series diverges if $\;|x-a| > R\;$

Now the above statement is False in my homework correction but I do not understand why?

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    Is it because of the <= sign in |x-a| <= R which I have read in my classe definition is only a < sign? Because the <= would mean it converges also for the boundaries of my convergence interval which is not necessarily true. Is that the expected reasoning? – 2017-01-14
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    @So It may depend on **what** were you asked: was it about the *radius of convergence* of a power series? Then yes, it is wrong for the weak inequaity, and perhaps it needs more explanation. – 2017-01-14
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    The series may or may not converge on the boundary. – 2017-01-14

2 Answers 2

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Take this simple on for example:

$$\sum_{n=0}^\infty x^n$$

it has radius of convergence $R=1$ by the ratio test, but, at $x=-1$, we get

$$\sum_{n=0}^\infty(-1)^n$$

which doesn't even pass the term test. The correct modified statement is that

Converges: $|x-a|

Unknown: $|x-a|=R$

Diverges: $|x-a|>R$

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On the circle $|x-a|=R$, it is possible for the series to converge on some points and not converge on others. Wikipedia has examples relating to convergence on the boundary.