Hints? Convergence behaviour.

Question

I tried many things, including ratio test, root test, comparison test, cauchy criterion, rewriting w.r.t e^n and doing all these tests again, but no luck in any direction. Any hints?

It seems most likely that I missed out on rewriting this stuff to resemble something that can be compared.

The root test definitely works here. – leoli1 Feb 14 '21 at 15:47 — leoli1, Feb 14 '21 at 15:47

score 1 · Answer 1 · answered Feb 14 '21 at 15:48

1

Root test should work since $$\sqrt[n]{\frac{n^{\ln n}}{(\ln n)^n}} = \frac{n^{\frac{\ln n}{n}}}{\ln n} = \frac{e^\frac{\ln^2n}{n}}{\ln n}$$

answered Feb 14 '21 at 15:48

VIVID

11,604

This is simpler than my approach. $+1$. My approach has the benefit of giving a rate of convergence, so it isn’t useless. – Clayton Feb 14 '21 at 15:49

score 0 · Answer 2 · answered Feb 14 '21 at 15:46

Hint: Rewrite the summand so that the numerator and denominator are written as exponentials (that is, $n^{\ln n}=\exp(\ln^2(n))$. Now you can combine the numerator’s and denominator’s exponents since they have the same base. You can show the resulting exponent is greater than $n$ (for example), so you can make a comparison to $\sum e^{-n}$.

Hints? Convergence behaviour.

2 Answers2