= 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 proves
$$
S_n=\int_n^{n+1}f(x)\,dx<f(n)<\int_{n-1}^nf(x)\,dx=S_{n-1}.
$$
Summing the rectangle bounds gives
$$
\int_2^{M+1}f(x)\,dx<\sum_{n=2}^Mf(n)<f(2)+\int_2^Mf(x)\,dx.
$$
The <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 $n=N+1,N+2,\ldots$ to obtain
$$
\int_{N+1}^\infty f(x)\,dx<R_N:=\sum_{n=N+1}^\infty f(n)<\int_N^\infty f(x)\,dx.
$$
For $f(x)=1/(x\ln x)$, the <antiderivative> is $\ln(\ln x)$, which grows without bound, so the first <series> \b[diverges]. For $f(x)=1/[x(\ln x)^2]$, the <antiderivative> is $-1/\ln x$, whose tail <integral> is finite, so the second <series> \b[converges], with
$$
\boxed{\frac1{\ln(N+1)}<R_N<\frac1{\ln N},\qquad N\geq2.}
$$
Both <functions> are positive and strictly decreasing on $[2,\infty)$, as required.
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