# Find whether an Alternating Sequence is Convergent

$$\sum _{n=2}^{\infty} \frac{\left(-1\right)^{n-1}}{\left(\ln \left(n\right)\right)^k} \quad \text{ for } k>0$$

By Leibniz's test if $$a_n \to 0$$ for an alternating sequence it is convergent, over here if I take the value of $$k\rightarrow 0$$ and $$\lim _{n\to \infty }\left(a_n\right)$$ then I would be able to say it is convergent. However, the limit appears to tend to $$0$$ when $$k \to 0$$ implying that the sequence is divergent. Am I right in this approach or am I doing something wrong while finding the limit. Is this series truly divergent?

• $k$ is a fixed positive number. The question of letting $k \to 0$ does not arise. Sep 7, 2023 at 7:58
• On an intuitive level the significance of $~k~$ approaching $~0,~$ from above is that as $~k~$ decreases, one can informally say that the alternating series takes longer to converge. Sep 7, 2023 at 8:46

The sequence has a term $$a_n = \frac{\left(-1\right)^{n-1}}{\left(\ln \left(n\right)\right)^k } \to 0$$ when $$n \to \infty$$, decreases in absolute value, and the sequence is alternating, so its sum converges.
(The proof of this classical result is: as $$a_n$$ decreases in absolute value and has alternating sign, $$\forall N, \forall k > N$$, the sum from $$2$$ to $$k$$ is bounded by the sums from $$2$$ to $$N$$ and from $$2$$ to $$N+1$$, an interval whose length is $$|a_{N+1}|$$ which $$\to 0$$ when $$N \to \infty$$. So the sequence of sums to $$N$$ is a Cauchy sequence, in a compact set (bounded and complete), so it converges).
You are confused by the fact that there is a parameter $$k$$. But $$k$$ is fixed for a given sequence, so you do not have to think about what happens when $$k \to 0$$.
• For the alternating series criterion, it's important that $|a_n|$ decreases as well as converges to $0$ (consider a series whose terms alternate between $1/n$ and $-1/n^2$). Sep 7, 2023 at 8:43