Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / Section B / 9

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.

The ratio test uses $L={\lim }_{n\to \infty }|a_{n+1}/a_n|$, when it exists for eventually nonzero terms: $L<1$ gives absolute convergence, $L>1$ gives divergence, and $L=1$ is inconclusive.

For $a_n=n/(2^n-1)$,Thus the first series converges absolutely. For the alternating second series,so it diverges. Alternating signs cannot rescue terms whose absolute values fail to tend to zero.