Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 1 / 9D / a / Solution

Suppose $a_n\to L$. Given $\epsilon >0$, choose $N$ such that $|a_k-L|<\epsilon /2$ for $k>N$. ThenThe first term tends to zero because its numerator is fixed, and the second is at most $\epsilon /2$. Thus the Cesaro mean tends to $L$.

The converse is false. For $a_n=(-1{)}^n$, the Cesaro means tend to $0$, while the original sequence alternates between $-1$ and $1$ and does not converge.

Solved by gpt-5.6-sol high.