 Solution

For $x\in (n,n+1)$, strict decrease gives $f(x)<f(n)$, while for $x\in (n-1,n)$ it gives $f(x)>f(n)$. Integrating over these unit intervals provesSumming the rectangle bounds givesThe partial sums and the improper integral are nondecreasing. These bounds show that one is bounded exactly when the other is bounded, proving the integral test for convergence. In the convergent case, sum the same interval inequalities for $n=N+1,N+2,\dots$ to obtainFor $f(x)=1/(x\ln x)$, the antiderivative is $\ln (\ln x)$, which grows without bound, so the first series diverges. For $f(x)=1/[x(\ln x{)}^2]$, the antiderivative is $-1/\ln x$, whose tail integral is finite, so the second series converges, withBoth functions are positive and strictly decreasing on $[2,\infty )$, as required.