My lecturer said it converges at $x = -3$ and diverges at $x=3$. I agree it diverges at $x = 3$ because it is the series $\sum\frac{3}{n}$. But at $x = -3$, isn't it $\sum-\frac{3}{n}$ which diverges as well?
2026-04-13 04:41:36.1776055296
Bumbble Comm
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How does the series $\sum\frac{x^{n+1}}{n3^{n}}$ converge when $x = -3$?
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Bumbble Comm
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No, the series is $$\sum_{n\ge1} \dfrac{3(-1)^{n+1}}{n}$$ which converges to $3 \log2$. (See this.)
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At $x=-3$ we get the series
$$3 \sum \frac{(-1)^{n+1}}{n}.$$
Greetings from Leibniz !