I'm trying to prove that for a sequence $(a_n)$ of reals, we have $$\limsup_{n \to \infty} \frac{1}{\log n} \sum_{k = 1}^n \frac{a_k}{k} \leq \limsup_{n \to \infty} \frac{1}{n} \sum_{k = 1}^n a_k.$$
My attempt was to use $$\sum_{i = 1}^n \frac{1}{i} = \log n + \gamma + O \left( \frac{1}{n} \right),$$ as $n \to \infty$, but I didn't really see how I could get rid of the $a_k$'s.
Can anyone help me on that?
Thanks!
Let $L = \limsup_{n\to \infty}\frac{1}{n}\sum_{k=1}^n a_k.$ Suppose $L\in \mathbb R.$ Let $\epsilon >0.$ Then there is $n_0\in \mathbb N$ such that for $n>n_0,$
$$\tag 1 \frac{1}{n}\sum_{k=1}^{n} a_k <L+\epsilon.$$
For $k=0,1,\dots,$ set $S_0=0, S_k = a_1 + \cdots + a_k, k = 1,2,\dots.$ Summing by parts, we have
$$\tag 2\sum_{k=1}^{n}\frac{a_k}{k} = \sum_{k=1}^{n}\frac{S_k-S_{k-1}}{k} = \frac{S_n}{n} + \sum_{k=1}^{n-1} \frac{S_k}{k(k+1)}.$$
For $n>n_0,$ $(1)$ then gives
$$\tag 3 \sum_{k=1}^{n} \frac{a_k}{k} = \frac{S_n}{n} + \sum_{k=1}^{n_0} \frac{S_k}{k(k+1)}+\sum_{k=n_0+1}^{n-1} \frac{S_k}{k(k+1)}$$ $$ \le L+\epsilon + \sum_{k=1}^{n_0} \frac{S_k}{k(k+1)}+(L+\epsilon)\cdot \sum_{k=n_0+1}^{n-1} \frac{1}{k+1}.$$
Now the last sum on the right is asyptotic to $\ln n$ as $n\to \infty.$ So dividing by $\ln n$ in $(3)$ and letting $n\to \infty$ gives
$$\tag 4 \limsup \frac{1}{\ln n}\sum_{k=1}^{n} \frac{a_k}{k} \le L+\epsilon.$$
Since $\epsilon$ was arbitrary, the left side of $(4)$ is $\le L$ as desired.
In the above we assumed $L\in \mathbb R.$ The case $L=\infty$ is obvious. I'll leave the case $L=-\infty$ to the reader for now.