Is the following series convergent? Let $A=\sum_{1}^{\infty}a_{n}$ be a convergent series. For every $n$, define
$$b_{n}=\frac{a_{1}+2a_{2}+...+na_{n}}{n(n+1)}.$$ 
Is $\sum_{1}^{\infty}b_{n}$ convergent?
 A: Rearrange this sum, then we get
$$\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty \frac{1}{n(n+1)}\sum_{k=1}^n k a_k = \sum_{n=1}^\infty  na_n \left( \frac{1}{n(n+1)}+\frac{1}{(n+1)(n+2)}+\cdots\right)=\sum_{n=1}^\infty a_n.$$

Strict proof: We consider partial sum of $\sum b_n$. Partial sum of $\sum b_n$ is
$$\frac{1}{N+1}\sum_{n=1}^N (N+1-n) a_n.$$
Let $\sum a_n =L$, then
$$\left| \frac{1}{N+1}\sum_{n=1}^N (N+1-n) a_n-L  \right|\le \frac{|L|}{N+1}+\frac{1}{N+1} \sum_{n=1}^N |a_1+a_2+\cdots +a_n -L|$$
By def. of limit, for all $\varepsilon>0$ there exists $M$ such that $n>M$ then $|a_1+a_2+\cdots +a_n -L|<\varepsilon$. Suppose $N>M$, then
$$\sum_{n=1}^N |a_1+a_2+\cdots +a_n -L| \le \sum_{n=1}^M |a_1+a_2+\cdots +a_n -L|+\sum_{n=M+1}^N \varepsilon$$
So
$$\left| \frac{1}{N+1}\sum_{n=1}^N (N+1-n) a_n-L  \right|\le \frac{|L|}{N+1}+\frac{1}{N+1}  \sum_{n=1}^M |a_1+a_2+\cdots +a_n -L|+\frac{N+1-M}{N+1}\varepsilon$$
take $N\to\infty$ then
$$\lim_{N\to\infty}\left| \frac{1}{N+1}\sum_{n=1}^N (N+1-n) a_n-L  \right|\le \varepsilon$$
for all $\varepsilon>0$. So $\sum b_n=L$.
