Prove that $y_n=\frac{x_1}{1^b}+\frac{x_2}{2^b}+... +\frac{x_n}{n^b} $ is convergent Let $a\ge 0$ and $(x_n) _{n\ge 1}$ be a sequence of real numbers. Prove that if the sequence $\left(\frac{x_1+x_2+...+x_n}{n^a} \right)_{n\ge 1}$ is bounded, then the sequence $(y_n) _{n\ge 1}$, $y_n=\frac{x_1}{1^b}+\frac{x_2}{2^b}+... +\frac{x_n}{n^b} $ is convergent $\forall b>a$.
To me, $y_n$ is reminiscent of the p-Harmonic series, but I don't know if this is actually true. Anyway,  I think that we may use the Stolz-Cesaro lemma on $\frac{x_1+x_2+...+x_n}{n^a}$, but I don't know if this is of any use. 

Let's do an Abel transformation. Denoting by $S_k = x_1 + ... + x_k$ for all $k \in \mathbb{N}$, you have
$$\sum_{n=1}^N \frac{x_n}{n^b} = x_1 +\sum_{n=1}^{N-1} \frac{S_{n+1}-S_n}{(n+1)^b} = x_1 +\sum_{n=1}^{N-1} \frac{S_{n+1}}{(n+1)^b} - \sum_{n=1}^{N-1} \frac{S_n}{(n+1)^b} $$
$$= x_1  +\sum_{n=2}^{N} \frac{S_n}{n^b} - \sum_{n=1}^{N-1} \frac{S_n}{(n+1)^b} =x_1 + \frac{S_N}{N^b} - \frac{x_1}{2^b} + \sum_{n=2}^{N-1} S_n \left( \frac{1}{n^b}-\frac{1}{(n+1)^b}\right) $$
Now, by hypothesis, $S_N = O(N^a)$, so because $b > a$, you have $S_N = o(N^b)$. So you have only to study the convergence of the series 
$$\sum_{n=2}^{N-1} S_n \left( \frac{1}{n^b}-\frac{1}{(n+1)^b}\right) $$
But one has $S_n = O(n^a)$, so 
$$S_n \left( \frac{1}{n^b}-\frac{1}{(n+1)^b}\right) = O \left( \frac{n^a}{n^b}-\frac{n^a}{(n+1)^b}\right)$$
Moreover, $$\frac{n^a}{n^b}-\frac{n^a}{(n+1)^b} = \frac{n^a(n+1)^b - n^an^b}{n^b(n+1)^b} \sim \frac{n^a}{(n+1)^b} \left( \left(1+ \frac{1}{n} \right)^b-1\right)$$
$$ \sim \frac{n^a}{(n+1)^b} \left( \frac{b}{n} \right) \sim \frac{b}{n^{b-a+1}}$$
This last series converges, so you deduce that the series of general term
$$ S_n \left( \frac{1}{n^b}-\frac{1}{(n+1)^b}\right)$$
also converges, so you deduce that your original series converges.
