# Power series and range of convergence

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.

• In your work, what was your reasoning re placing $(n+1)$ in the denominator? – user2661923 Aug 16 '20 at 20:32
• The ratio test tells you the radius of convergence=7, but that means convergence in the OPEN interval (-7,7). – herb steinberg Aug 16 '20 at 21:42

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.