# Does $\sum\limits_{k=1}^{n}\frac{(-1)^k}{k}(-1)^n$ converge?

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?

• This isn't a series, it's just a partial sum. Nov 7 '20 at 19:34
• @Invisible you're right, but it's clear what he is asking :) Nov 7 '20 at 19:34
• @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 $\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{(-1)^k}{k}(-1)^{n}$? Nov 7 '20 at 19:40
• $\sum\limits_{n=1}^\infty a_n$ is a conventional notation for series. Nov 14 '20 at 7:52

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