<s>[INST] Range of Convergence of $\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \ 3^n (x-5)^n}$ $$\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \ 3^n (x-5)^n}$$
I am trying to use the alternating series test to find a range of $x$ for which $(1) b_n > b_{n+1}$ and $ (2) \lim_{n \to \infty} \frac{1}{n \ 3^n (x-5)^n} = 0$. If $|\frac{1}{x-5}| \leq 1$ then condition $(1)$ and $(2)$ will not hold. So wouldn't the range be $x < 4$ and $5 \leq x$ ? I know this is not right since the answer should be $ 5\frac{1}{3} \leq x$ and $x < 4 \frac{2}{3}$ ... could someone provide a solution?
 [/INST] Hints : by Cauchy Hadamard test , range will be $ |x-5|\gt\frac{1}{3} \implies x\gt \frac{16}{3},x\lt \frac{14}{3} $.
Only equality $x=\frac{16}{3} $ is possible because series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} $ is converge.
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