Consider the power series
[tex]\displaystyle\sum_{n\ge1}\dfrac{x^n}n[/tex]
By the ratio test, this series converges for
[tex]\displaystyle\lim_{n\to\infty}\left|\frac{x^{n+1}}{n+1}\cdot\frac n{x^n}\right|=|x|\lim_{n\to\infty}\frac n{n+1}=|x|<1[/tex]
though we know by the alternating series test that the series converges for [tex]x=-1[/tex].
So this series converges for [tex]-1\le x<1[/tex].
Differentiating the series yields
[tex]\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n\ge1}\frac{x^n}n=\sum_{n\ge1}x^{n-1}=\sum_{n\ge0}x^n[/tex]
which is the geometric series. We know this series converges for [tex]|x|<1[/tex], and this time the endpoints are not included.
This example shows that (A) is certainly possible; that is, [tex]x=-1[/tex] is valid in the first series, but not in the differentiated one.
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Now consider the series
[tex]\displaystyle\sum_{n\ge0}\dfrac{x^n}{n!}[/tex]
which we know to converge to [tex]e^x[/tex].
Differentiating, we get
[tex]\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n\ge1}\frac{x^{n-1}}{(n-1)!}=\sum_{n\ge0}\frac{x^n}{n!}=e^x[/tex]
as expected. But both series converge everywhere, so this serves as a counter-example to the claim of B. So B is false.
