Root test to test for convergence or divergence

Question

infinity
∑	(1/nln(n))^n
n=2

Answer 1

_n=2∑^∞(1/nln(n))ⁿ.

(1) L=limit as n→∞ (1/(nln(n)))ⁿ or (2) L=limit as n→∞ ((1/n)ln(n))ⁿ?

(1) nln(n)→∞ as n→∞, so (nln(n))ⁿ→∞ as n→∞, and its reciprocal → 0, so L=0. This is less than 1 so the series is convergent.

(2) ((1/n)ln(n))ⁿ=(1/nⁿ)lnⁿ(n). Let n=10 (for example) then this becomes 10^-10ln¹⁰(10). In this example ln¹⁰(10)<<10¹⁰, so the limit as n→∞ is 0, making L=0 implying a convergent series.

In each case the series is convergent using the root test.

Since n=2 is the lower limit ln(n)>0 and n>0, no individual terms are indeterminate (that is, none tend to infinity).