Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 1 / 11E (Analysis I)

For positive $a_n,b_n$,The comparison test for series therefore gives

The arithmetic-geometric mean inequality givesThus convergence of ${\sum }_n{a}_n$ implies convergence of ${\sum }_n\sqrt{a_n{a}_{n+1}}$ by the comparison test for series.

The converse is false. DefineThe even terms make ${\sum }_n{a}_n$ diverge, whileso the neighboring geometric-mean series converges.

Suppose ${\sum }_n{a}_n$ diverges. Positivity implies $s_n\to \infty$. Put $t_n=a_n/s_n\in (0,1)$ for $n\ge 2$. Since $s_{n-1}=s_n-a_n$,If ${\sum }_n{t}_n$ converged, then eventually $t_n\le 1/2$ andSumming would bound the telescoping quantity $\log (s_n/s_1)$, contradicting $s_n\to \infty$. ThereforeThe converse is true. If ${\sum }_n{a}_n<\infty$, then $s_n\ge s_1>0$ andTaking the contrapositive shows that divergence of the ratio series forces divergence of ${\sum }_n{a}_n$.