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$$\sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \ 3^n (x-5)^n}$$

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?

A. Radek Martinez
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    Try different values of x: $4$, $4 \frac 23$ (I'm sure it should be $\frac 23$), $4 \frac 56$, $5 \frac 16$, $5 \frac 13$, $6$. For each case, it should be pretty clear why it converges/diverges. The general answer can be derived from them. Also, after you consider $4$, it should be clear why $4 \frac 23$ is the next number to consider. – Dmitry Aug 07 '20 at 01:44

3 Answers3

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What about the ratio test? \begin{align*} \limsup_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right| & = \limsup_{n\to\infty}\frac{n3^{n}|x-5|^{n}}{(n+1)3^{n+1}|x-5|^{n+1}}\\\\ & = \limsup_{n\to\infty}\left(\frac{n}{n+1}\right)\frac{1}{3|x-5|} = \frac{1}{3|x-5|} < 1\\\\ & \Rightarrow |x - 5| > \frac{1}{3} \Rightarrow \left(x > \frac{16}{3}\right)\vee\left(x < \frac{14}{3}\right) \end{align*}

Moreover, for $x = 16/3$, one has that \begin{align*} \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^{n}3^{-n}} = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} < \infty \end{align*} due to the Leibniz test.

On the other hand, for $x = 14/3$, we have that \begin{align*} \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n3^{n}(-1)^{n}3^{-n}} = -\sum_{n=1}^{\infty}\frac{1}{n} = \infty \end{align*} since it is the harmonic series.

Hopefully this helps.

user0102
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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.

A learner
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By the root test, the series converses for all $x$ such that $$ \limsup_n\sqrt[n]{\frac{1}{n3^n|x-5|^n}}=\frac{1}{3|x-5|}\lim_n\frac{1}{\sqrt[n]{n}}=\frac{1}{3|x-5|}<1 $$

Thus, the series converges for all $x$ such that $|x-5|>\frac{1}{3}$, i.e., all $x$ in $(-\infty,\tfrac{14}{3})\cup(\tfrac{15}{3},\infty)$.

At the point $x=\frac{15}{3}$ the series becomes $\sum_n\frac{(-1)^{n-1}}{n}$ which converges (is an alternating series with decreasing driving term $\frac{1}{n}$). At $x=\frac{14}{3}$ the series becomes $-\sum_n\frac{1}{n}$ which diverges. Therefore, the series converges for all $x\in (-\infty,\tfrac{14}{3})\cup[\tfrac{15}{3},\infty)$

Mittens
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