Convergence of the series $\sum_{n=1}^{\infty}\frac{x_n}{n^b}$ 
Assume the sequence $y_n=\dfrac{x_1+x_2+\cdots+x_n}{n^a}$ is bounded, where $a>0$ and $x_n$ is a sequence of real numbers. Prove that if $b>a$ then  $\sum\limits_{n=1}^{\infty}\dfrac{x_n}{n^b}$ is convergent. 

I have tried to solve in the case a=1, b=2 by calculating x with respect to y, we get 
$$\sum\limits_{n=1}^{\infty}\dfrac{x_n}{n^b}=\sum\limits_{n=1}^{\infty}\dfrac{x_n}{n^2}=\dfrac34 y_1+\cdots+\dfrac{2n+1}{(n+1)^2}y_n+\cdots$$
However, even in this simple case I cannot get the conclusion. Could you please help me with this? Thank you very much for your help.
 A: A slight modification of Dirichlet's test for convergence works for this problem:
Let $s_0=0$ and $s_n=x_1+\dots+x_n$ for $n\geq 1$, and choose a constant $C$ such that $|s_n|\leq Cn^a$ for all $n$. Then 
$$ \sum_{n=1}^N\frac{x_n}{n^b}=\sum_{n=1}^N\frac{s_n-s_{n-1}}{n^b}=\frac{S_N}{N^b}+\sum_{n=1}^{N-1}s_n[n^{-b}-(n+1)^{-b}] $$
by summation by parts.
For the first term, we have
$$ \Big|\frac{S_N}{N^b}\Big|\leq \frac{C}{N^{b-a}}\to 0$$
as $N\to\infty$.
For the sum, since $n^{-b}>(n+1)^{-b}$ it follows that
$$ \sum_{n=1}^{N-1}|s_n(n^{-b}-(n+1)^{-b})|=\sum_{n=1}^{N-1}|s_n|(n^{-b}-(n+1)^{-b})\leq C\sum_{n=1}^Nn^a(n^{-b}-(n+1)^{-b})=$$
$$=Cb\sum_{n=1}^{N-1}n^a\int_{n}^{n+1}x^{-b-1}\;dx\leq Cb\sum_{n=1}^{N-1}\int_n^{n+1}x^{-(b-a)-1}\;dx=Cb\int_1^{N}x^{-(b-a)-1}\;dx$$
Taking $N\to\infty$ and using the fact that $(b-a)+1>1$, we see that
$$ \sum_{n=1}^{\infty}|s_n(n^{-b}-(n+1)^{-b})| $$
converges, and hence that
$$ \sum_{n=1}^{\infty}s_n(n^{-b}-(n+1)^{-b}) $$
converges. Therefore
$$ \sum_{n=1}^{\infty}\frac{x_n}{n^b}=\sum_{n=1}^{\infty}s_n(n^{-b}-(n+1)^{-b})$$
converges as well.
A: Try the integral vertion, and see if you can imitate the steps, i.e.: show that if $$ \frac{1}{u^a} \int_1^u X(t) \space dt $$
is bounded, then 
$$ \int_1^{+\infty} \frac{X(t)}{t^b} \space dt $$
is convergent for any $b > a$.
A: I will elavorate on my early suggestion. It was meant to lead in the direction of @carmichael561 solution.  
Suppose $ C = \sup_{u \ge 1} \frac{1}{u^a} |\int_1^u X(t) \space dt | \tag{1} $ is finite.  
Then, to relate $\int_1^{+\infty} \frac{X(t)}{t^b} \space dt$ to  (1), consider how to involve $ S(u) = \int_1^u X(t) \space dt $. Since $S'(u)=X(u)$ we have:
$$ 
\int_1^{+\infty} \frac{X(t)}{t^b} \space dt = 
\int_1^{+\infty} \frac{S'(t)}{t^b} \space dt =
 \frac{A(t)}{t^b}{\LARGE |}_1^{\infty} - \int_1^{+\infty} S(t)\frac{(-b)}{t^{b+1}} \space dt. \tag{2}
$$
Next use the hypothesis that $|S(t)|/t^a$ is bounded to conclude that both terms in the last expression in (2) are finite.  
What does that have to do with your problem? Well, integrals are limits of sums, so may be what worked for integrals works for your question. Also, in the case of monotone functions, integrals and sums are related (one can bound sums with integrals and viceversa). Let's replace integrals in the previous argument and let's see what happens.
We introduce $S(n) = \sum_{t=1}^n x_t$ (here $t$ takes values in $\mathbb N$, I am using the same symbol as in the integral case to underlie similarities). What is the relation between $S$ and $x$? In the continuous case it involves derivatives, in the discrete case it involves differences: $x_n=S(n) - S(n-1)$ (with $S(0)=0$). Then we proceed as @carmichael561 :
$$
\sum_{n=1}^N\frac{x_n}{n^b}=
\sum_{n=1}^N\frac{S(n)-S(n-1)}{n^b}=
\frac{S(N)}{N^b}+\sum_{n=1}^{N-1}S(n)[n^{-b}-(n+1)^{-b}] .\tag{3}
$$
In the continuous case, instead of the difference $n^{-b}-(n+1)^{-b}$, we had the derivative of $\varphi(t) = 1/t^b$. But the difference between two values of $1/t^b$ can be related to a value of its derivative as in the intermediate value theorem:
$$ n^{-b}-(n+1)^{-b} = -( \varphi(n+1) - \varphi(n) ) = - \varphi'(n+\theta) = \frac{b}{(n+\theta)^{b+1}} \le \frac{b}{n^{b+1}} \tag{4}
$$ 
for some $\theta \in (0,1)$ . Here $\theta$ depends on $n$ and $b$, it's exact value is not important, we know it is between $0$ and $1$, so we know an approximate value of $\theta$, and what we know is enough to justify the last inequality in (4).  
So the terms in the last series in (3) satisfy:
$$
|S(n)| \large{(} \frac{1}{n^b} - \frac{1}{(n+1)^b} \large{)} \le 
|S(n)| \frac{b}{n^{b+1}} \le
C \frac{b}{n^{b-a+1}}.
$$
Therefore the series is summable.
