School and Work

Note that, by taking the limit of the n-th term as n --> infinity:
lim (n-->infinity) ln(n + 1)/ln(n^3 + 2)
= lim (n-->infinity) ln[n(1 + 1/n)]/ln[n^3(1 + 2/n^2)]
= lim (n-->infinity) [ln(n) + ln(1 + 1/n)]/[3ln(n) + ln(1 + 2/n^2)]
= lim (n-->infinity) [1 + ln(1 + 1/n)/ln(n)]/[3 + ln(1 + 2/n^2)/ln(n)]
= (1 + 0)/(3 + 0)
= 1/3.

Since this limit does not converge to zero, this series is divergent by the limit test.

I hope this helps!

View the original article here