Is the following series convergent?

Question

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?

Answer 1

Rearrange this sum, then we get $$\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty \frac{1}{n(n+1)}\sum_{k=1}^n k a_k = \sum_{n=1}^\infty na_n \left( \frac{1}{n(n+1)}+\frac{1}{(n+1)(n+2)}+\cdots\right)=\sum_{n=1}^\infty a_n.$$

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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 $\sum b_n=L$.