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Check if the following series diverge or converge: $$\sum_\limits{n=1}^{\infty}(-1)^n\frac{\ln n}{n}$$.

My Solution: I have tried to apply Leibniz' criterion $\lim_{n\to\infty}(-1)^n\frac{\ln n}{n}=0$

I need also to check if $\frac{\ln n+1}{n+1}<\frac{\ln n}{n}$. I computed the derivative: $d\frac{\frac{\ln n}{n}}{n}=\frac{1-\ln(n)}{n^2}$, which is decreasing, so the Leibniz criterion can be applied and series converge.

The book solution: "converges conditionally"

Question:

What is conditional convergence? Is my attempt right?

Thanks in advance!

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Yes you are right since

$$f(x)=\frac{\ln x}x\implies f’(x)=\frac{1-\ln x}{x^2}$$

which is negative for $x>e$ and thus $\ln n/n\to 0$ is strictly decreasing and the series converges by Leibniz.

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What is conditional convergence? Is my attempt right?

I'll be answering this since it seems to me that you don't know what condition convergence is, rather than solving the question.

First of all, your attempt is correct. The series is convergent. As for what is a conditional convergence, we say that $\sum a_k$ converges conditionally if $$\lim\limits_{n\to \infty}\sum_{k=1}^{n} a_k = L$$ While at the same time, $$\lim\limits_{n\to \infty}\sum_{k=1}^{n} |a_k| = + \infty$$ Or alternately, the limit of the absolute values does not converge.

In your question, we have that $a_k = (-1)^k \frac{\ln k}{k}$ and you proved that the series converges, while $|a_k|$ is $\frac{\ln k}{k}$ and you can take it as an exercise to prove that $\lim\limits_{n\to \infty}\sum_{k=1}^{n} |a_k| = \infty$

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A slightly less formal way to think about conditional convergence is to ask the question:

If those alternating signs weren't there, would this series converge?

Without the alternating signs, the series would be $\sum \frac{\ln n}{n}$, which definitely diverges by the comparison test. On the other hand with the alternating signs the series does converge, because the individual terms go to zero. So in a sense it's the alternating signs that "save" convergence here. That's what we mean by saying the convergence is "conditional". (If a series converges either way, i.e. with or without the absolute values, then the convergence is "absolute".)

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A serie is conditionally convergent if it is convergent, but is not absolutely convergent.

This is the case here as the series of absolute values $\sum_{n\ge 1}\dfrac{\log n}n$ diverges, by the comparison with the (divergent) harmonic series.

Your proof is correct, but I think it's better not to use derivatives. It's easy to show the sequence $\dfrac{\log n}n$ is decreasing: you only have to prove the difference of two consecutive terms is $>0$: $$\frac{\log n}n-\frac{\log (n+1)}{n+1}=\frac{(n+1)\log n-n\log(n+1)}{n(n+1)}$$ so we only have to prove that $\; (n+1)\log n-n\log(n+1)>0$. \begin{align}\text{Now }\qquad(n+1)\log n-n\log(n+1)&=(n+1)\log n-n\log n-n\log\Bigl(1+\frac1n\Bigr)\\ &=\log n-n\log\Bigl(1+\frac1n\Bigr)\begin{aligned}[t] &\ge\log n-n\,\frac 1n=\log n-1\\&(\text{as }\; \log(1+u)\le u),\end{aligned} \end{align} which is $>0$ for all $n\ge 3$.

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Given $\{a_n\}_{n\in \mathbb N}\subset \mathbb R$. The series $\sum_{n=1}^\infty a_n$ conditionally converges if $\sum_{n=1}^\infty a_n$ converges but $\sum_{n=1}^\infty |a_n|$ diverges.

Since

$$ \left| (-1)^n \frac{\ln n}{n}\right|\geq \frac{1}{n}, \ \text{for} \ n\geq 3 $$ and the harmonic series diverges, we conclude that $\sum (-1)^n \frac{\ln n}{n}$ conditionally converges.

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    $\begingroup$ I think you skipped convergence. $\endgroup$ – djechlin Jun 22 '18 at 17:39
  • $\begingroup$ he already did that $\endgroup$ – Gonzalo Benavides Jun 22 '18 at 17:48
  • $\begingroup$ I actually thought the writing in the question was unclear but I see what you mean. It's better to make your answer self-contained since people more likely come here for answers than questions. You can include "The OP already showed converge of the original series" if you don't want to repeat the argument. $\endgroup$ – djechlin Jun 24 '18 at 16:56

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