Find all values of $x \in \mathbb R$ for which the series $\sum_{n=1}^\infty \frac{(-1)^nx^n}{n}$ is $(i)$ absolutely convergent, $(ii)$ convergent in the usual sense.

My solutions: (can someone tell me if I am correct?) Is the radius of convergence the same for absolute and normal convergence, so we only have to check the endpoints separately in both cases?

$(i)$ Apply the ratio test for power series to the sequence $a_n$ := $\frac{(-1)^n}{n}$. We have $R = $$\lim_{n\to \infty} |\frac{a_{n+1}}{a_{n}}| = \lim_{n\to \infty}\frac{n} {n+1}$ $=1$. So the series definitely converges for $x \in (-1,1)$.

Check endpoints: $x=1$: $\sum_{n=1}^\infty \frac{1}{n}$ diverges as harmonic series.

$x=-1$: $\sum_{n=1}^\infty \frac{1}{n}$ diverges by harmonic series.

So $x \in (-1,1)$.

$(ii)$ The radius of convergence is $1$ so definitely converges for $x \in (-1,1)$.

$x=1$: $\sum_{n=1}^\infty \frac{(-1)^n}{n}$ converges by alternating series test.

$x=-1$: $\sum_{n=1}^\infty \frac{1}{n}$ diverges by linearity of summation and harmonic series.

So for conditional convergence: $x \in (-1,1]$.

  • 1
    $\begingroup$ The radii of convergence and absolute convergence of the series $\sum a_nx^n$ coincide as they are equal to $\sup\{r\,:\, a_nx^n\to 0,\ |x|< r\}.$ Observe that $a_nx^n\to 0$ iff $|a_nx^n|\to 0.$ $\endgroup$ Commented Jan 17, 2023 at 1:47

2 Answers 2


We consider the more general problem: For what complex values of $z$ does the series $$\sum_{n=1}^\infty \frac{(-1)^n z^n}{n}$$ converge? The concept of radius of convergence comes arises naturally in the two-dimensional complex plane.

The radius of convergence is $R = 1/\alpha$, where $ \alpha = \limsup_{n\rightarrow\infty}\sqrt[n]{|c_n|}$, where the power series is written $\sum c_n z^n$. In our case $R = 1$ (where $\limsup$ is also equal to the ordinary limit). The root test asserts that the series converges for $|z|< R$, and diverges for $|z| > R$, but says nothing about the points on the circle $|z| =R$. Note that if $|z|< R$ ($|z|>R$), then the series converges (diverges) both normally and absolutely. The radius of convergence is the only place where we may have one but not the other, but there it tells us nothing about either case, and other methods must be used, as you note. There is only one radius of convergence; there isn't one for absolute convergence and another for (ordinary) convergence.

Your solution is correct. More generally, the series can be shown to converge at all complex values with $|z|\le 1$ except $z = -1$. See, e.g., Rudin, Principles of Mathematical Analysis, Theorem 3.44. This series, by the way, is known as the Mercator (or Newton-Mercator) series, and can be obtained as the Taylor expansion of $-\log(1+z)$; note that at $z=-1$ the logarithm is infinite.


Your solution is correct.

Yes, it is correct that for a power series, the radius of absolute convergence is the same as the radius of convergence.

More specifically, for a power series,

  • if the radius of convergence is $0$, then it does not converge absolutely anywhere except at $0$
  • if the radius of the convergence is positive, then it converges absolutely inside the open disk of radius equal to the radius of convergence. Moreover, it does not converge anywhere outside the closed disk of radius equal to the radius of convergence. (If we are only concerned about real variable, replace "disk" by "interval".)

So, yes, you only have to check the endpoints of the interval separately in both cases, since whether it is convergent or absolute convergent is the same anywhere else.

I prefer to avoid speaking "the radius of normal convergence". The meaning of "the radius of convergence" is pretty much conventional and standard. There is no need to emphasize its "normal-ness", even in this context of comparison. An extra adjective might introduce some unintentional or unnecessary connotation.

  • $\begingroup$ @APassJack To confirm, is my solution correct ? $\endgroup$
    – user1071088
    Commented Jan 17, 2023 at 9:51
  • 1
    $\begingroup$ @NikitaMazepin Yes. Also see my updated answer. $\endgroup$
    – Apass.Jack
    Commented Jan 17, 2023 at 10:04

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