Question

A.If a new series is found by differentiating a known series with included endpoints, those endpoints might not be included in the new series. 
B.If a new series is found by differentiating a known series with included endpoints, those endpoints will never be included in the new series.
which statements are false?

Answer 1

Consider the power series

$\displaystyle\sum_{n\ge1}\dfrac{x^n}n$

By the ratio test, this series converges for

$\displaystyle\lim_{n\to\infty}\left|\frac{x^{n+1}}{n+1}\cdot\frac n{x^n}\right|=|x|\lim_{n\to\infty}\frac n{n+1}=|x|$

though we know by the alternating series test that the series converges for $x=-1$.

So this series converges for $-1\le x$.

Differentiating the series yields

$\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n\ge1}\frac{x^n}n=\sum_{n\ge1}x^{n-1}=\sum_{n\ge0}x^n$

which is the geometric series. We know this series converges for $|x|$, and this time the endpoints are not included.

This example shows that (A) is certainly possible; that is, $x=-1$ is valid in the first series, but not in the differentiated one.

- - -

Now consider the series

$\displaystyle\sum_{n\ge0}\dfrac{x^n}{n!}$

which we know to converge to $e^x$.

Differentiating, we get

$\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n\ge1}\frac{x^{n-1}}{(n-1)!}=\sum_{n\ge0}\frac{x^n}{n!}=e^x$

as expected. But both series converge everywhere, so this serves as a counter-example to the claim of B. So B is false.