So the integral test says that if the integral of the series converges, then the series converges as well. To integrate this, start with the substitution u=ln(n) -> . So the numerator is simply u^2 and from the du term we have (1/n)dn There is still an extra n unaccounted for though in the denominator. So notice that e^u=n by our original substitution. Thus we can write that this integral is also and it seems you're allowed to simply state that this converges.