I am trying to determine whether the series $\displaystyle\sum_{n=2}^{+\infty} \ln\left(1-\frac{(-1)^n}{\sqrt n}\right)$ converges or not.

I have tried using a Taylor series of the summand, which gives $$v_n = -\frac{(-1)^n}{\sqrt n} - \frac{1}{2n} + O\left(\frac{1}{n\sqrt n}\right)$$ whose series does not converge because the first term gives an alternating series, the big-O is absolutely convergent but the middle term is the harmonic series, and is thus equivalent to $-\frac{1}{2}\ln n$. Thus the series diverges and the partial sum tends to negative infinity.

However, Wolfram Alpha tells me the exponential of the series, the infinite product $\displaystyle\prod_{n=2}^{+\infty}\left(1-\frac{(-1)^n}{\sqrt n}\right)$ has a nonzero limit, which contradicts that.

Have I been doing something wrong ?

  • $\begingroup$ Maybe you mistyped the Wolfram Alpha input. The series indeed diverges to $-\infty$. Something you definitely did wrong was to write $\frac{1}{2k}$ instead of $\frac{1}{2n}$ in the expansion of $v_n$. $\endgroup$ – Daniel Fischer Oct 4 '15 at 20:17
  • $\begingroup$ Well, the product of two consecutive terms of the form $1-\frac{(-1)^n}{\sqrt{n}}$ is $1-\frac{1}{n}+O\left(\frac{1}{n\sqrt{n}}\right)$, and your proof of divergence looks perfectly fine. $\endgroup$ – Jack D'Aurizio Oct 4 '15 at 20:22
  • $\begingroup$ @DanielFischer: Wolfram|Alpha does give a non-zero limit. $\endgroup$ – joriki Oct 4 '15 at 20:42
  • $\begingroup$ @DanielFischer Whoops. I copied straight over from a sheet of paper where it was $v_k$. Thank you I edited the original post. $\endgroup$ – Groovy. Oct 4 '15 at 21:02
  • $\begingroup$ Good, so @joriki's link proves that Wolfram Alpha gives a wrong answer. While that doesn't positively prove that you didn't mistype, we can relegate that to a merely theoretical possibility and conclude that you haven't done anything wrong here. The error is on the other side of the internet connection. $\endgroup$ – Daniel Fischer Oct 4 '15 at 21:07

Wolfram Alpha was wrong, and my proof was correct. We indeed have $$ \prod_{n=2}^\infty \left(1-\frac{(-1)^n}{\sqrt n}\right) = 0 $$ as $\displaystyle\sum _{k=2}^n \ln\left(1-\frac{(-1)^k}{\sqrt k}\right) = -\frac{1}{2}\ln(n) + \mathrm O(1)$ as $n $ approaches infinity.

  • $\begingroup$ interesting path that could have been taken is to take: $$\exp\sum_{k=2}^n \ln\left(1-\frac{(-1)^k}{\sqrt k}\right)) = \prod_{n=2}^{\infty}e^{ln(1-\frac{(-1)^{n}}{\sqrt(n)}}$$ Now all one, has to show: $$\lim_{N \rightarrow \infty}\prod_{n=2}^{N}e^{ln(1-\frac{(-1)^{n}}{\sqrt(n)}} = 0$$ $\endgroup$ – Zophikel Dec 22 '17 at 23:45
  • $\begingroup$ I think this can be done via a Stirling's approximation $\endgroup$ – Zophikel Dec 22 '17 at 23:45

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