Does $\sum_{n=1}^{\infty} \frac{(-3/2)^n}{n}$ diverge or converge?

By the nth term test, since $\lim _{n \to \infty} \frac{(-3/2)^n}{n} \neq 0,$ I get that the series diverges.

However, I also have seen that $\sum_{n=1}^{\infty} \frac{x^n}n = \ln\big(\frac1{1-x}\big),$ so with $x=-\frac32,$ I thought the series would converge conditionally.

What am I doing wrong?

  • 6
    $\begingroup$ You are right, the series diverges. $\sum_{n=1}^{\infty} \frac{x^n}n = \ln\big(\frac1{1-x}\big)$ only applies when $|x| < 1$ or when $x = -1$. $\endgroup$
    – Bruno B
    Jun 19 at 17:16
  • 2
    $\begingroup$ @StevenClark Why? $\endgroup$ Jun 19 at 17:21

1 Answer 1


The equality $\sum_{n=1}^\infty\frac{x^n}n=\log\left(\frac1{1-x}\right)$ holds if and only if $x\in[-1,1)$. In particular, it does not hold when $x=-\frac32$.

  • $\begingroup$ Thank you! Does [-1,1) come from the ratio test and the fact that we cannot have x=1 ($\ln0$)? $\endgroup$
    – ada
    Jun 19 at 17:30
  • 3
    $\begingroup$ @ada yes, take the power series and find its radius of convergence. then check the end points $\endgroup$ Jun 19 at 17:41

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