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Given that $\sum_{n=1}^{\infty}a_nx^n$ is a power series that its range of convergence is $[-7,7]$, I need to determine if the following statements is true:

The power series $\sum_{n=1}^{\infty}na_nx^{n-1}$ converges at $[-7,7]$.

I found that the radius of convergence of this power series is also 7:

$$\lim\limits_{n \to \infty} \big| \frac{na_n}{(n+1)a_{n+1}}\big|=\lim\limits_{n \to \infty} \big| \frac{n}{n+1}\big|\big| \frac{a_n}{a_{n+1}}\big|=1\cdot7=7$$

The answer is that this is false, but I don't know why and how to disprove it.

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  • $\begingroup$ In your work, what was your reasoning re placing $(n+1)$ in the denominator? $\endgroup$ – user2661923 Aug 16 '20 at 20:32
  • $\begingroup$ The ratio test tells you the radius of convergence=7, but that means convergence in the OPEN interval (-7,7). $\endgroup$ – herb steinberg Aug 16 '20 at 21:42
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Consider the example $$a_n = \begin{cases} \frac{(-1)^{n / 2}7^{-n}}n, &\text{ if }2\mid n;\\0, &\text{ else}.\end{cases}$$ That is, the original series is $$\sum_{m = 1}^\infty\frac{(-1)^m (x / 7)^{2m}}{2m}.$$ It is easy to see that this series converges (not necessarily absolutely) for any $x\in[-7, 7]$. But the power series $\sum_{n = 1}^\infty na_nx^{n - 1}$ does not converge at $x = \pm7$, as the general term does not tend to zero.

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