Here's what I came up with:
For this problem, I'm required to use a comparison test to determine if $\Sigma1/ln(n)^n$ converges or diverges. By intuition, I am thinking that $\Sigma1/ln(n)^n$ converges.
To prove that it converges by the Direct Comparison Test, I would have to find a convergent series of a sequence that is greater than $1/ln(n)^n$ for $n > \epsilon$. I find that the sequences that I know do not satisfy this condition.
$x^n$, $1/n^n$, $1/n^p$, $1/n!$ each converge, but are not greater than $1/ln(n)^n$.
So then I move on to prove convergence by the Limit Comparison Test. Each of the infinite series of the sequences that I know converge all fail the conditions required by the Limit Comparison Test, since the limit of the ratio of the two ends up being $0$ or $infinity$, and now I am stuck.
What sequence I should compare $1/ln(n)^n$ to?
$\endgroup$ 12 Answers
$\begingroup$Well, for $n > e^2$, $\ln n > 2$, so $(\ln n)^n > 2^n$ whenever $n > e^2$, and thus $$\sum_{n = 9}^\infty \frac{1}{(\ln n)^n} < \sum_{n = 9}^\infty \frac{1}{2^n} < \sum_{n = 0}^\infty \frac{1}{2^n} = 2$$ since $e^2 < 3^2 = 9$ and $\sum_{n = 0}^\infty 1/2^n$ is a geometric series. Therefore, $\sum_{n = 9}^\infty 1/(\ln n)^n$ converges, and $$\sum_{n = 1}^\infty \frac{1}{(\ln n)^n} = \sum_{n = 1}^8 \frac{1}{(\ln n)^n} + \sum_{n = 9}^\infty \frac{1}{(\ln n)^n}$$ also converges because $\sum_{n = 1}^8 1/(\ln n)^n$ is finite.
$\endgroup$ $\begingroup$Observe that, for $N$ sufficiently large, $\ln n > 2 \iff \dfrac{1}{(\ln n)^n} < \dfrac{1}{2^n}$ for all $n\geq N$. So you can compare this with a convergent GP and obtain the result.
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