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Let $a_k\geq 0$ be a non-increasing sequence satisfying $\lim_{k\to\infty} a_k = 0$. Is it always true that $$\lim_{n\to\infty} \frac{\sum_{k=1}^n a_k}n=0$$ ?

I tried proving with "brute force" but failed: Let $\varepsilon>0$. Then there exists a $K\in\mathbb N$ such that $a_k\le\varepsilon$ for all $k\geq K$. Let $C=\sum_{k=1}^K a_k$. We have $$\frac{\sum_{k=1}^{\left\lceil\frac C\varepsilon\right\rceil}a_k}{\left\lceil\frac C\varepsilon\right\rceil}=\frac{C}{\left\lceil\frac C\varepsilon\right\rceil}+\sum_{k=K+1}^{\left\lceil\frac C\varepsilon\right\rceil} a_k\le \varepsilon +\varepsilon\cdot\left( \left\lceil\frac C\varepsilon\right\rceil-K\right).$$

However, this is not good, as the right-hand side depends on $C$, which can get large for small $\varepsilon$.

Also, numerical computations for example with $a_k=\frac1{\log\log k}$ seem to confirm the statement though...

(Additionally, note that I didn't use the non-increasing fact, but I am not sure if it is needed.)

ArtOfProblemSolving
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2 Answers2

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You're on the right track. But you don't need to take $n = \lceil\frac{C}{\epsilon}\rceil$. Simply take $n>K+1$. Then you have:$$ \frac{\sum_{k=1}^n a_k}{n} = \frac{\sum_{k=1}^K a_k}{n} + \frac{\sum_{k=K+1}^n a_k}n \le \frac{\sum_{k=1}^K a_1}{n} + \frac{\sum_{k=K+1}^n \epsilon}n \le \frac{K a_1}{n} + \frac{(n-K)\epsilon}n $$ As $n$ goes to infinity, the RHS approaches $\epsilon$. In particular, for $n>N$ (with appropriate choice of $N$), the whole thing is bounded by $2\epsilon$.

Dark Malthorp
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Continue from "Then there is a $K\gt 0$ such that $a_i\lt\epsilon$ for all $i\gt K$.

(Say the sequence started from $a_0$)
Then let $N$ be an integer larger than $\max\{\frac {a_0} {\epsilon} K,K\}$. Since $a_i$ is non increasing, $\sum _{i=0} ^K a_i \le Ka_0$, and $\sum _{i=K} ^N a_i\le (N-K)\epsilon$. Then for all $n\gt N$, $\frac{\sum_{k=1}^n a_k}n\lt\epsilon$.

user12986714
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