Answer:
Converges
Step-by-step explanation:
n = k+1
(k+1) ÷ [ln(k+1)]^(k+1)
n = k
k ÷ [ln(k)]^k
{(k+1) ÷ [ln(k+1)]^(k+1)} ÷ {k ÷ [ln(k)]^k}
= {(k+1)/k} × {[ln(k)]^k}/{[ln(k+1)]^(k+1)}
{[ln(k)]^k}/{[ln(k+1)]^(k+1)} is decreasing at a much faster rate than the rate of increase of {(k+1)/k}.
Therefore the product is less than 1.
Hence it converges
