Does $\sum_{i=1}^{\infty} i^{-p}a_i$ converge for some $p>1$ given that $\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{i=1}^{n}a_i$ converges? Given a non-negative sequence $a_i\geq 0$ and $\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{i=1}^{n}a_i = a^* < \infty$. Can we show that, for some $p>1$,
$$ \lim_{n\rightarrow\infty} \sum_{i=1}^{n} i^{-p}a_i < \infty$$
I have tried using the one-sided limit comparison test as follows
$$\lim\sup_{i\rightarrow\infty} \frac{i^{-p}a_i}{n^{-1}a_i} \leq \lim\sup_{i\rightarrow\infty}\frac{i}{i^{p}} = \lim\sup_{i\rightarrow\infty} i^{1-p}=1$$
Is the above proof correct? I am tempted to conclude that my conjecture is correct and the second sequence converges. But what puzzles me is the value $n$ in the limit comparison test. 
Thanks in advance! 
 A: In fact, if $\lim_{n \to \infty} \frac{1}{n}\sum_{i=1}^n a_i$ converges, then $\sum_{i=1}^\infty i^{-p} a_i$ converges for any $p > 1$.
Every convergent sequence is bounded above, so there exists $M > 0$ such that for all $n$, $\frac{1}{n}\sum_{i=1}^n a_i \leq M$. So $\sum_{i=1}^n a_i \leq Mn$ for all $n$.  
Let $S_n = \sum_{i=1}^n i^{-p} a_i$, the sequence of partial sums of $\sum_{i=1}^\infty i^{-p} a_i$.
Consider the series $\sum_{i=1}^\infty i^{-p} M$, which converges by the p-series test for $p > 1$. The partial sums of this sequence must be a convergent sequence, and therefore must be bounded. Set the partial sums as $T_n = \sum_{i=1}^n i^{-p} M$; thus there exists $K> 0$ such that for all $n$, $T_n < K$.
We will show that $S_n \leq T_n$ for all $n$, which makes $S_n < K$ for all $n$. This implies $S_n$ is bounded above and as it is monotonically increasing, it must converge.
Fix $N$, and consider all finite sequences $\{a_i\}_{i=1}^N$ of length $N$ such that $\sum_{i=1}^n a_i \leq Mn$ for all $n \leq N$. Suppose $\{a_i\}$ is one of these sequences that maximizes the value of $S_N$. We can show that this sequence must be $a_i = M$ for all $i$. 
First, suppose that $a_j < M$ for some $j$. Since $\sum_{i=1}^{j-1} a_i \leq M(j-1)$, then $\sum_{i=1}^{j} a_i < Mj$. It could be that $a_k = 0$ for all $k > j$, in which case we can clearly increase $a_j$ to $M$ while still satisfying $\sum_{i=1}^n a_i \leq Mn$ for all $n \leq N$. This will make a sequence with a larger value of $S_N$. Otherwise there exists $a_k > 0$ for some $k > j$. Then we can decrease $a_k$ by a small amount (for example some $\epsilon$ where $\epsilon  < M - a_j$, and $\epsilon < a_k$) and increase $a_j$ by the same amount. Since $j < k$, $$(a_j + \epsilon)j^{-p} + (a_k - \epsilon)k^{-p} > (a_j)j^{-p} + (a_k)k^{-p}$$ so this will again create a satisfactory sequence with a larger value of $S_N$. 
Suppose instead that $a_j > M$ for some $j$. We know $\sum_{i=1}^{j} a_i \leq Mj$, and if $a_i \geq M$ for all $i < j$, we would have  $\sum_{i=1}^{j} a_i > Mj$, so it must be $a_i < M$ for some $i < j$. In this case the previous argument applies for $a_i$.
Therefore, the sequence that maximizes $S_N$ must have $a_i = M$ for all $i$, and for this sequence, $S_N = T_N$. This finishes our proof that $S_n \leq T_n$ for all possible sequences $\{a_i\}$ satisfying our condition.
Since the partial sums $S_n$ are a convergent sequence, by definition the sum  $\sum_{i=1}^\infty i^{-p} a_i$ converges for any $p > 1$.
I would not doubt there is a simpler proof of this.
A: We have $\sum_{i=1}^na_i\le M\,n$ for some $M>0$. Using Abel' summation formula we get
$$\begin{align}
\sum_{i=1}^na_i\,i^{-p}&=\Bigl(\sum_{i=1}^na_i\Bigr)n^{-p}+\sum_{i=1}^{n-1}\Bigl(\sum_{j=1}^ia_j\Bigr)(i^{-p}-(i+1)^{-p}))\\
&\le M\,n^{1-p}+M\sum_{i=1}^{n-1}i\,(i^{-p}-(i+1)^{-p})\\
&\le M+M\,p\sum_{i=1}^{\infty}i^{-p}.
\end{align}$$
Since the partial sums are bunded and the terms are positive, the series converges.
A: It is easy to see that any $p > 2$ works. We have $$0 \le \frac{a_n}{n} \le \frac{a_1 + \dotsb + a_n}{n} := \frac{s_n}{n} \to a$$
so $a_n = O(n)$, and then $$\sum_{n=1}^\infty \frac{a_n}{n^p} = \sum_{n=1}^\infty \frac{O(1)}{n^{p-1}}$$
which is finite for $p-1 > 1$. Here I'm using $O(1)$ to mean some eventually bounded quantity. But I suspect more is possible, 
as the condition $s_n/n \to a$ actually implies $a_n = o(n)$, as the following identity shows:
$$\frac{a_n}{n} = \frac{s_n}{n} - \frac{n-1}{n}\frac{s_{n-1}}{n-1} \to a - a = 0.$$
Of course, the best possible result would be convergence for all $p > 1$; taking $a_n = a$ shows that $p \le 1$ cannot work.
