- #1

tmt1

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

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?