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).