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.

Printable View

- Mar 7th 2008, 01:07 PMnatesterProof 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.

- Mar 8th 2008, 12:26 AMCaptainBlack
$\displaystyle b_N=\frac{1}{N}~\sum_{n=1}^N a_n$

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

$\displaystyle b_N=\frac{1}{N}~\left[\sum_{n=1}^M a_n+\sum_{n=M+1}^N a_n\right]$ $\displaystyle

\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 $\displaystyle 0$ as $\displaystyle N$ goes to $\displaystyle \infty$.

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

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

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

RonL - Mar 8th 2008, 02:58 PMThePerfectHacker
If fact is $\displaystyle a_n \to a$ then $\displaystyle b_n \to a$ too. Where $\displaystyle b_n$ is an arithmetic sequence.

- Mar 9th 2008, 01:20 AMCaptainBlack