MHB Finding the interval of convergence for a series with lnn

[tmt1]

So I have

$$\sum_{n = 2}^{\infty} \frac{1}{nln(n)}$$

I'm trying to apply the limit comparison test, so I can compare it to $b_n$ or $\frac{1}{n}$ and I can let $a_n = \frac{1}{nln(n)}$

Then I get $$\lim_{{n}\to{\infty}} \frac{n}{nln(n)}$$

Or $$\lim_{{n}\to{\infty}} \frac{1}{ln(n)}$$ Which is clearly 0.

Now, since the limit is 0, then the sums of series of $\sum_{ }^{}a_n$ would only converge if $\sum_{}^{}b_n$ converges. However $\sum_{}^{}b_n$ equals $\sum_{}^{}\frac{1}{n}$ which diverges.

The answer is that it is divergent, but I'm not sure how to prove it with my method. Is the method I am using not the right one to use, or am I misunderstanding the method?

 

[Prove It]

So I have

$$\sum_{n = 2}^{\infty} \frac{1}{nln(n)}$$

I'm trying to apply the limit comparison test, so I can compare it to $b_n$ or $\frac{1}{n}$ and I can let $a_n = \frac{1}{nln(n)}$

Then I get $$\lim_{{n}\to{\infty}} \frac{n}{nln(n)}$$

Or $$\lim_{{n}\to{\infty}} \frac{1}{ln(n)}$$ Which is clearly 0.

Now, since the limit is 0, then the sums of series of $\sum_{ }^{}a_n$ would only converge if $\sum_{}^{}b_n$ converges. However $\sum_{}^{}b_n$ equals $\sum_{}^{}\frac{1}{n}$ which diverges.

The answer is that it is divergent, but I'm not sure how to prove it with my method. Is the method I am using not the right one to use, or am I misunderstanding the method?

I would use the integral test.

First let $\displaystyle \begin{align*} f(x) = \frac{1}{x\ln{(x)}} \end{align*}$. We should note that $\displaystyle \begin{align*} f'(x) = -\frac{\left[ \ln{(x)} + 1 \right] }{\left[ x \ln{(x)} \right] ^2 } \end{align*}$ which is negative for all $\displaystyle \begin{align*} x \geq 2 \end{align*}$, and thus $\displaystyle \begin{align*} \frac{1}{x\ln{(x)}} \end{align*}$ is decreasing in this region.

As it's decreasing, that means $\displaystyle \begin{align*} \sum_{n = 2}^{\infty} \frac{1}{n\ln{(n)}} > \int_2^{\infty}{ \frac{1}{x\ln{(x)}}\,\mathrm{d}x } \end{align*}$, so evaluating the integral we have

$\displaystyle \begin{align*} \int_2^{\infty}{ \frac{1}{x\ln{(x)}}\,\mathrm{d}x } &= \int_{\ln{(2)}}^{\infty}{ \frac{1}{u}\,\mathrm{d}u } \textrm{ after substituting } u = \ln{(x)} \implies \mathrm{d}u = \frac{1}{x}\,\mathrm{d}x \\ &= \lim_{\epsilon \to \infty} \left[ \ln{|u|} \right] _{\ln{(2)}}^{\epsilon} \\ &= \lim_{\epsilon \to \infty} \ln{(\epsilon )} - \ln{\left[ \ln{(2)}\right] } \\ &\to \infty \end{align*}$

As the sum is greater than this infinite value, the sum also diverges.