0
$\begingroup$

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

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?

$\endgroup$
2
  • 5
    $\begingroup$ $k$ is a fixed positive number. The question of letting $k \to 0$ does not arise. $\endgroup$ Sep 7, 2023 at 7:58
  • 1
    $\begingroup$ 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. $\endgroup$ Sep 7, 2023 at 8:46

1 Answer 1

1
$\begingroup$

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 $k \to 0$.

By the way, when talking about a sequence whose successive terms are summed, the convention is to call it a series rather than sequence.

$\endgroup$
2
  • $\begingroup$ 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 $-1/n^2$). $\endgroup$
    – ronno
    Sep 7, 2023 at 8:43
  • $\begingroup$ @ronno Yes, thanks for noting that I forgot this condition. $\endgroup$ Sep 7, 2023 at 8:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .