Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 1 / 9F (Analysis I)

For a positive-term series, the ratio test says that if $a_{n+1}/a_n\to L<1$, the series converges; if the ratio tends to $L>1$, including positive infinity, it diverges. More generally $\mathrm{limsup}{a}_{n+1}/a_n<1$ suffices for convergence and $\mathrm{liminf}{a}_{n+1}/a_n>1$ for divergence. A limiting ratio of one gives no conclusion.

The P-series $\sum n^{-2}$ converges and the harmonic series $\sum n^{-1}$ diverges. Both have strictly decreasing positive terms, and their successive ratios, $(n/(n+1){)}^2$ and $n/(n+1)$, tend to one. These supply the two requested possibilities.

The function $1/x$ is decreasing and positive. Therefore on $[k,k+1]$ it is at most $1/k$, and the Riemann integral givesSumming from $k=1$ to $n-1$ yieldsFor $n=1$ both sides are zero; for $n>1$ the inequality is actually strict. The same integral test for convergence gives ${\sum }_{k=N}^{n-1}1/k\ge \log (n/N)$.

First, for $0<x<1$, the derivative of $h(x)=\log (1-x)+x$ is $-x/(1-x)<0$, while $h(0)=0$. Hence $\log (1-x)<-x$.

Choose $N>c$ large enough that the given ratio bound holds for every $n\ge N$. Positivity permits taking logarithms, and the preceding inequality givesSumming, and using the integral test for convergence from part (b), gives for $n>N$Thus $a_n\le a_N{N}^c{n}^{-c}$. Since $c>1$, the P-series $\sum n^{-c}$ converges, so the comparison test for series provesThe finitely many initial terms do not affect convergence. This power-law comparison from a refined ratio bound supplies information when the ordinary ratio test can be inconclusive.