Does the alternating series converge? [Leibniz criterion]

Question

Does the series $$\sum_{k=1}^\infty(-1)^{k-1}\frac{\ln k}{k}$$converge?

What I have tried: It does converge. I applied the Leibniz criterion

1)$$\frac{\ln k}{k}\geq0 \forall k\in\mathbb{N}$$ so the series is indeed alternating.

2)I showed that $$\frac{\ln (k+1)}{k+1}<\frac{\ln k}{k},$$when $k>2$ by introducing function $f(x)=\frac{\ln x}{x}$ and showing that $f'(x)<0,$ when $x>e$.

3) $$\lim_{k\to\infty}\frac{\ln k}{k}=0.$$ The question is: can I ignore the first two terms in part 2) as $$\frac{\ln (k+1)}{k+1}<\frac{\ln k}{k},$$ only when $k>2$?

@Atmos: sources differ. Some have $0\in\mathbb{N}$ and others have $0\not\in\mathbb{N}$. — robjohn, Feb 15 '18 at 16:59
If you change sequence in only finitely many terms, the series of the first sequence converges if and only if the series of the second sequence converges. — Thomas Andrews, Feb 15 '18 at 17:23
@robjohn Really ? For the non zero set it is common to use $\mathbb{N}^{*}$ or am I wrong ? — Atmos, Feb 15 '18 at 17:32
It is convergent by Leibniz' test and convergent to $\frac{1}{2}\log^2(2)-\gamma\log(2)$ by the reflection formula for the $\zeta$ or $\eta$ function. — Jack D'Aurizio, Feb 15 '18 at 18:41
@JackD'Aurizio: this is shown in equation $(8)$ of this answer. — robjohn, Feb 15 '18 at 19:21

score 3 · Answer 1 · answered Feb 15 '18 at 16:52

3

Yes you can "ignore" it. You split out this two terms $$ \sum_{k=1}^{+\infty}\left(-1\right)^{k-1}\frac{\ln\left(k\right)}{k}=0-\frac{\ln(2)}{2}+\sum_{k=3}^{+\infty}\left(-1\right)^{k-1}\frac{\ln\left(k\right)}{k} $$ So what you've done is good.

answered Feb 15 '18 at 16:52

Atmos

7,369

Does the alternating series converge? [Leibniz criterion]

1 Answers1