1. ## Proof analysis

If {a_n} is a null sequence, and for each n, b_n is the arithmetic mean of the first n terms of {a_n}, then prove {b_n} is a null sequence.

2. Originally Posted by natester
If {a_n} is a null sequence, and for each n, b_n is the arithmetic mean of the first n terms of {a_n}, then prove {b_n} is a null sequence.
$b_N=\frac{1}{N}~\sum_{n=1}^N a_n$

Put $M=\lfloor \sqrt{N} \rfloor$ then:

$b_N=\frac{1}{N}~\left[\sum_{n=1}^M a_n+\sum_{n=M+1}^N a_n\right]$ $
\le \frac{{\rm{sup}}_{n\in [1,M]}(a_n)}{\sqrt{N}}+\frac{N-(\sqrt{N}-1)}{N}{\rm{sup}}_{n \in [M,N]}(a_n)
$

The first term on the right clearly goes to $0$ as $N$ goes to $\infty$.

The coefficient of the second term has a finite limit as $N$ goes to $\infty$, but because $\{a_n\}$ is a null sequence:

$\lim_{N \to \infty}{\rm{sup}}_{n \in [M,N]}(a_n)=0$

which proves that $\{b_n\}$ is a null sequence

RonL

3. If fact is $a_n \to a$ then $b_n \to a$ too. Where $b_n$ is an arithmetic sequence.

4. Originally Posted by ThePerfectHacker
If fact is $a_n \to a$ then $b_n \to a$ too. Where $b_n$ is an arithmetic sequence.
Which is a corrolary of what the original poster was asked to prove

RonL