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Classify $$\sum_{n=3}^\infty \frac{1}{n\log(n)}$$ as absolutely convergent, conditionally convergent or divergent.

Is it, $$\sum_{n=3}^\infty \frac{1}n$$ is a divergent $p$-series as $p=1$, and $$\lim_{n\to\infty} \frac{1}{n\log(n)}{n} = 0$$ by comparison test. And this converges to $0$.

So,
$$\sum_{n=3}^\infty \frac{1}{n\log (n)}$$ is conditionally convergence?

I'm not sure if I'm doing right or not. Could you guide me?

Thanks in advance! :)

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  • $\begingroup$ A sequence can only be conditionally convergent if the terms in the series alternate between positive and negative. So we could say that $\sum_{n=1}^{\infty} \frac{(-1)^n}{n \log(n)}$ converges conditionally, but the sequence as you presented in your question diverges. $\endgroup$ – Omnomnomnom Jun 11 '13 at 3:51
  • $\begingroup$ You can use the main result. $\endgroup$ – Mhenni Benghorbal Jun 11 '13 at 4:13
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Note that if your given series is convergent then it's also absolutely convergent since $\frac{1}{n\log n}\geq 0\quad \forall n\geq 3$.

Now, since the sequence $(\frac{1}{n\log n})$ is decreasing to $0$ so by the integral test your series has the same nature that the improper integral $$\int_3^\infty\frac{dx}{x\log x}=\left[\log(\log(x))\right]_3^{\to\infty}=\infty$$ hence the series is divergent.

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  • $\begingroup$ By definition, a series cannot be both conditionally convergent and absolutely convergent - conditionally convergent means "convergent, but not absolutely conergent". See, for instance, the wikipedia entry $\endgroup$ – Nick Peterson Jun 11 '13 at 4:20
  • $\begingroup$ @nrpeterson Thanks for your precious comment. $\endgroup$ – user63181 Jun 11 '13 at 7:03
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There are a couple of issues here. First and foremost, this sum can't start with $n=1$... when $n=1$, you are dividing by $\log n=0$. Beyond that:

1) Remember that conditional convergence means that the series converges, but that the series obtained by replacing each term by its absolute value does not. Since this series only has positive terms, the series of absolute values is exactly the same - so, this series must either diverge or converge absolutely.

2) Please, review the comparison test; it is a matter of inequalities, not limits. There is a variant commonly called the limit comparison test, but it requires that the limit be 0. Besides, all that this limit being 0 establishes is that your series is smaller than a series that diverges... and being $\leq\infty$ doesn't tell us much!

For this particular series, the integral test is the way you want to go. Why? Because the function $f(x)=\frac{1}{x\log x}$ can be integrated nicely - remember, the derivative of $\log x$ is $\frac{1}{x}$, making a substitution appropriate.

So, there are a few steps here.

1) Show that $f(x)$ and the series satisfy the conditions of the integral test. Namely, it must be the case that $f(n)=a_n$ for all natural numbers $n$, where $a_n$ is the $n$th term of the series; that $f(x)\geq 0$ for all $x\geq 0$; and that $f(x)$ is decreasing. (In this case, it actually only decreases for $x\geq e$, but that's fine... why?)

2) Given this, the integral test tells us that $\sum_{n=2}^{\infty}\frac{1}{n\log n}$ and $\int_2^{\infty}\frac{dx}{x\log x}$ have the same convergence behavior. So, it is a matter now of establishing whether or not this improper integral converges.

I'll leave it to you to fill in the details!

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  • $\begingroup$ oh wow, thank you for the helpful notes and tips that I am careless and blindly without realizing at all.. I got it show divergent now. $\endgroup$ – Risa Jun 11 '13 at 4:08

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