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I'm trying to prove that, given $(u_n)_n \in \mathbb{C}^\mathbb{N}$ verifying $ u_{n+1}-u_n =_{n} o(\frac{1}{n})$, the following holds:

$$ \lim_{n\to\infty} \frac{u_1+...+u_n}{n} = a \in \mathbb{C} \implies \lim_{n\to\infty} u_n = a$$

It is the reciprocal to the Cesaro 'average theorem'.

An indication is given, to rewrite the average using $u_n$ and $a_k = u_k - u_{k-1}$ (with $a_1=u_1$) which I have figured out to be:

$$u_1+...+u_n = u_n + \sum\limits_{k=1}^n \sum\limits_{i=1}^k a_i$$

What I've tried doing is rewriting this double sum, which I've found to be a $o\big(ln(\small(n+1\small)!)\big)$. Other than that, most of my attempts have proven futile to solving this.

John Do
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1 Answers1

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Define $a_1=u_1$ and $a_n=u_{n}-u_{n-1}$ for $n\geq 2$.

Note that $$\begin{align} u_n-\frac{S_n}n &= u_n-\frac 1n \sum_{k=1}^nu_k \\ &=u_n-\frac 1n \sum_{k=1}^n \sum_{j=1}^k a_k \\ &=u_n-\frac 1n \sum_{j=1}^n\sum_{k=j}^n a_j\\ &=u_n-\frac 1n \sum_{j=1}^na_j\sum_{k=j}^n 1\\ &=\sum_{j=1}^n a_j-\frac 1n \sum_{j=1}^n (n-j+1)a_j\\ &=\frac{1}n\sum_{j=1}^n (j-1)a_j \end{align}$$

Since $na_n\to0$, $(n-1)a_n\to0$ and the usual Cesaro theorem yields $\lim_{n\to \infty} \frac{1}n\sum_{j=1}^n (j-1)a_j=0$.

This implies $\displaystyle \lim_n (u_n-\frac{S_n}n)=0$, hence $\lim_n u_n=a$.

Gabriel Romon
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