Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 1 / 10F (Analysis)

The integral test for convergence applies when $a_n=h(n)$ for all sufficiently large $n$, where $h$ is nonnegative, continuous and nonincreasing on $[N,\infty )$. It states that ${\sum }_{n=N}^{\infty }{a}_n$ converges if and only if the improper integral ${\int }_N^{\infty }h(x)dx$ is finite. Its comparison bounds areNo conclusion about an arbitrary nonnegative sequence follows without such a monotone comparison function; finite initial terms do not affect convergence.

For $h(x)=x^{-\alpha }$ with $\alpha >0$, all the hypotheses hold. If $\alpha \ne 1$,whereas for $\alpha =1$ the integral is $\log R$. The first expression has a finite limit exactly when $\alpha >1$. Thus the p-series satisfies

On $[3,\infty )$, $h(x)=1/(x\log x)$ is positive, continuous and decreasing, since its denominator has positive derivative $\log x+1$. Substituting $u=\log x$ givesThe integral test for convergence therefore shows that the series is divergent.

Because $3>e$, each of $x$, $\log x$ and $\log \log x$ is positive and increasing for $x\ge 3$. Hence $h(x)=1/[x\log x(\log \log x{)}^2]$ is positive, continuous and decreasing. With $u=\log \log x$, $du=dx/(x\log x)$, soThe integral test for convergence implies that the series is convergent.

Compare with the divergent series from part (i). The ratio of its new summand to $1/(n\log n)$ isFor completeness, $\log n={\int }_1^n{x}^{-1}dx\le {\int }_1^n{x}^{-1/2}dx=2(\sqrt{n}-1)$, so $(\log n)/n\to 0$. The limit comparison test now gives divergence. Although the exponent $1+1/n$ exceeds one for each summand, it approaches one and does not produce a convergent p-series with a fixed exponent greater than one.