$\sum_{n=1}^\infty \frac{a_n}{n^{1+\delta}}$ converges 
Let $\left\{a_n\right\}$ be real sequence,  $b_n=\frac{a_1+\cdots+a_n}{n}, n\geq 1$. Suppose that $\left\{b_n\right\}$ is bounded, show that for any $\delta>0$, the $\sum_{n=1}^\infty \frac{a_n}{n^{1+\delta}}$ converges.

My attempt: let $c_k=a_k/k$, then by $a_k=kb_k-(k-1)b_{k-1}$, we see readily that $$|\sum_{i=1}^n c_i|=|b_n+\sum_{k=2}^n \frac{b_{k-1}}{k}$$ But this could not prove that $|\sum_{i=1}^n c_i|$ is bounded, or else, we could use Dirichlet test... Any ideas? or other proof?

Since $nb_n-(n-1)b_{n-1}=a_n$, Abel's transform gives (denoting $B= \sup_{\ell\geqslant 1}\lvert b_\ell\rvert$)
\begin{align}
\left\lvert \sum_{n=N+1}^M\frac{a_n}{n^{1+\delta}}\right\rvert
&=\left\lvert \sum_{n=N+1}^M\frac{nb_n }{n^{1+\delta}}-\sum_{n=N+1}^M\frac{(n-1)b_{n-1} }{n^{1+\delta}}\right\rvert\\
&=\left\lvert \sum_{j=N+1}^M\frac{jb_j }{j^{1+\delta}}-\sum_{j=N}^{M-1}\frac{jb_{j} }{(j+1)^{1+\delta}}\right\rvert\\
&\leqslant \frac{\lvert b_M\rvert}{M^\delta}+\frac{\lvert b_N\rvert}{N^\delta}
+\left\lvert \sum_{j=N+1}^{M-1}jb_j\left(\frac{1 }{j^{1+\delta}}- \frac{1 }{(j+1)^{1+\delta}}\right)\right\rvert
\\
&\leqslant \frac{B}{M^\delta}+\frac{B}{N^\delta}
+B\sum_{j=N+1}^{M-1}j\left( \frac{1 }{j^{1+\delta}}- \frac{1 }{(j+1)^{1+\delta}} \right)
\end{align}
and the convergence of the wanted series follows from the convergence of $\sum_{j=1}^{\infty}j\left( \frac{1 }{j^{1+\delta}}- \frac{1 }{(j+1)^{1+\delta}} \right)$.
