Let's consider the series: $\sum\limits_{k=1}^{n}\frac{(-1)^k}{k}(-1)^n$.

I would suggest that it doesn't converge. One can see it as follows:

Let be $S_{2n}:=\sum\limits_{k=1}^{2n}\frac{(-1)^k}{k}(-1)^{2n}$ and $S_{2n+1}:=\sum\limits_{k=1}^{2n+1}\frac{(-1)^k}{k}(-1)^{2n+1}$ two subsequences (subseries?) of $\sum\limits_{k=1}^{n}\frac{(-1)^k}{k}(-1)^n$. Both converge due to alternating series test (Leibniz criterion). However, if I take a look at $|S_{2n}-S_{2n+1}|$, I notice:

$$|S_{2n}-S_{2n+1}|=\Big|\sum\limits_{k=1}^{2n}\frac{(-1)^k}{k}(-1)^{2n} -\sum\limits_{k=1}^{2n+1}\frac{(-1)^k}{k}(-1)^{2n+1}\Big|\\=\Big|2\sum\limits_{k=1}^{2n}\frac{(-1)^k}{k}(-1)^{2n} -\frac{1}{2n+1}\Big|=2\sum\limits_{k=1}^{2n}\frac{(-1)^{k-1}}{k} +\frac{1}{2n+1}>1. $$ Hence, $\sum\limits_{k=1}^{n}\frac{(-1)^k}{k}(-1)^n$ violates the Cauchy-criterion, or in other words I can always find two partial sums which are not arbitrarily close to each other.

Is this correct? Is there a faster or more elegant way two show this result?

  • 2
    $\begingroup$ This isn't a series, it's just a partial sum. $\endgroup$
    – Invisible
    Nov 7, 2020 at 19:34
  • $\begingroup$ @Invisible you're right, but it's clear what he is asking :) $\endgroup$ Nov 7, 2020 at 19:34
  • $\begingroup$ @Invisible Yes, I know what you are meaning but how do I write it correctly? I can't write something like $\sum\limits_{k=1}^{\infty}\frac{(-1)^k}{k}(-1)^{\infty}$, can I? Or maybe just $\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{(-1)^k}{k}(-1)^{n}$? $\endgroup$
    – Philipp
    Nov 7, 2020 at 19:40
  • $\begingroup$ $\sum\limits_{n=1}^\infty a_n$ is a conventional notation for series. $\endgroup$
    – Invisible
    Nov 14, 2020 at 7:52

2 Answers 2


The sequence $$u_n =\sum\limits_{k=1}^{n}\frac{(-1)^k}{k}$$

converges to $-\ln(2) < 0$. Therefore the sequence $$\sum\limits_{k=1}^{n}\frac{(-1)^k}{k}(-1)^n=(-1)^n u_n$$

does not converge.


Your proof is correct. $|S_{2n}-S_{2n+1}| > 1$ for all $n$ implies that $(S_n)$ is not a Cauchy sequence and therefore not convergent.

You also showed that $(S_{2n})$ and $(S_{2n+1})$ are convergent, but this is not needed for the proof.


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