Determine the series' convergence

(n^(ln n))/(ln n)^n

The use of powers suggests that we use the Root Test.

By the Root Test, if lim (n-->infinity) (a_n)^(1/n) < 1, then Σ a_n is convergent.

With a_n = n^[ln(n)]/[ln(n)]^n, we have:
lim (n-->infinity) (a_n)^(1/n) = lim (n-->infinity) {n^[ln(n)]/[ln(n)]^n}^(1/n)
= lim (n-->infinity) n^[ln(n)/n]/[ln(n)]^(n/n)
= lim (n-->infinity) n^[ln(n)/n]/ln(n).

Note that lim (n-->infinity) n^[ln(n)/n] is a limit that takes the form infinity^0 (lim (n-->infinity) ln(n)/n = 0 by L'Hopital's Rule). So, we need to compute it by letting:
L = lim (n-->infinity) n^[ln(n)/n].

Taking the natural logarithm of both sides:
ln(L) = lim (n-->infinity) ln{n^[ln(n)/n]}
= lim (n-->infinity) [ln(n)]^2/n
= 0.

Thus, L = e^0 = 1 and:
lim (n-->infinity) n^[ln(n)/n]/ln(n) = 0.

Therefore, the series converges.

I hope this helps!

Try the ratio test - I think you'll find the series consisting of these terms converges.