Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 1 / 10D (Analysis I)

Let $d_j=a_{j+1}-a_j$ and suppose $d_j\to L$. Telescoping givesTo justify the Cesaro mean limit directly, choose $J$ so that $|d_j-L|<\epsilon$ for $j\ge J$. The finite initial sum contributes at most $n^{-1}{\sum }_{j<J}|d_j-L|$, which tends to zero. The remaining terms contribute at most $\epsilon$, and replacing $(n-1)L/n$ by $L$ contributes $|L|/n$. Consequently $\boxed{a_n/n\to L}$.

Take $\boxed{a_n=(-1{)}^n}$. Then $|a_n/n|=1/n\to 0$, but $a_{n+1}-a_n=-2(-1{)}^n$ alternates between $2$ and $-2$. Thus convergence of the scaled sequence does not imply convergence of its successive differences.

No. Take $\boxed{a_n=\log n}$. For each fixed positive integer $k$,by continuity of the logarithm at $1$. Yet $a_n\to +\infty$, so $(a_n)$ is not a convergent sequence of real numbers. Vanishing differences at every fixed lag do not control the accumulation of many small increments.

Yes. The arbitrary variable-lag condition forces the Cauchy sequence property. Suppose otherwise. Then some ${\epsilon }_0>0$ has the property that, for every $N$, there exist $p,q\ge N$ with $|a_p-a_q|\ge {\epsilon }_0$. In particular, for each $n$, choose such $p,q\ge n$. By the triangle inequality, at least one satisfies $|a_p-a_n|\ge {\epsilon }_0/2$ or $|a_q-a_n|\ge {\epsilon }_0/2$. That index must be strictly greater than $n$, since the difference at $n$ itself is zero.

It follows that the following positive integer is defined for every $n$:For this particular function, $|a_{n+f(n)}-a_n|\ge {\epsilon }_0/2$ for every $n$, contradicting the assumed limit zero. Therefore $(a_n)$ is a Cauchy sequence, and the completeness of the real numbers proves it converges to a finite real limit. This is convergence from arbitrary variable-lag increments; the quantifier over every function is much stronger than the fixed-lag condition in the previous part.