Cesaro summablity implies the convergence of the Abel sum

Suppose $\Sigma c_n$ is cesaro summable to $0$, prove that for every $0\le r <1$ , $\Sigma c_n r^n$ is convergent.

(A series is cesaro summable if the sequence of the means of its partial sums converges)

If we know this , I know how to prove that Abel sum tends to $0$ ,when $r$ tends to $1$, and then the proof of "cesaro summablity implies Abel summablity " is complete .

Hint: Put $S_n=c_1+\cdots+c_n$. We have $\displaystyle \frac{S_1+\cdots+S_n}{n}=\frac{T_n}{n}=u_n\to 0$. Now $S_n=T_n-T_{n-1}=nu_n-(n-1)u_{n-1}$, and as $u_n\to 0$, there exists $M$ such that $|u_n|\leq M$ for all $n$. Then $|S_n|\leq M(2n-1)$. Now $S_n-S_{n-1}=c_n$, and we can bound $c_n$ easily.
• it is not true. $S_n / n$ doesn't tend to $0$.we just know that the mean of $S_n$ s tend to $0$. – user115608 Oct 23 '15 at 14:01
• yes we have a bound that depends on $n$ for $c_n$. But how to prove the convergence? – user115608 Oct 23 '15 at 16:17
• If you have $|c_n|\leq Kn$ for $n\geq 1$, $K$ constant, you need only show that the series $\sum n r^n$ is convergent. Put $a_n=nr^n$, you have $\frac{a_{n+1}}{a_n}\to r<1$, hence the convergence. – Kelenner Oct 23 '15 at 16:21