# Another Convergence Problem

• Nov 15th 2008, 01:06 PM
RedBarchetta
Another Convergence Problem
Let $\displaystyle \sum\limits_{n = 1}^\infty {a_n }$ be a convergent series of positive terms.

What can be said about the convergence of $\displaystyle \sum\limits_{n = 1}^\infty {\frac{{a_1 + a_2 + \cdots + a_n }} {n}}$ ?

My Attempt:

$\displaystyle \because \sum\limits_{n = 1}^\infty {a_n } = s_n \therefore \sum\limits_{n = 1}^\infty {\frac{{a_1 + a_2 + \cdots + a_n }} {n}} = \sum\limits_{n = 1}^\infty {\frac{{s_n }} {n}} = s_n \sum\limits_{n = 1}^\infty {\frac{1} {n}}$

Therefore, $\displaystyle \sum\limits_{n = 1}^\infty {\frac{{a_1 + a_2 + \cdots + a_n }} {n}}$ is a divergent harmonic series?
• Nov 15th 2008, 01:17 PM
flyingsquirrel
Hi,
Quote:

Originally Posted by RedBarchetta
$\displaystyle \sum\limits_{n = 1}^\infty {\frac{{s_n }} {n}} = s_n \sum\limits_{n = 1}^\infty {\frac{1} {n}}$

(Surprised) You can't do this, $\displaystyle s_n$ depends on $\displaystyle n$ !

Hint : for all $\displaystyle n\geq 1$, $\displaystyle s_n\geq a_1>0$ because $\displaystyle a_k>0$ for all $\displaystyle k\geq 1$.
• Nov 15th 2008, 01:20 PM
Mathstud28
Quote:

Originally Posted by RedBarchetta
Let $\displaystyle \sum\limits_{n = 1}^\infty {a_n }$ be a convergent series of positive terms.

What can be said about the convergence of $\displaystyle \sum\limits_{n = 1}^\infty {\frac{{a_1 + a_2 + \cdots + a_n }} {n}}$ ?

My Attempt:

$\displaystyle \because \sum\limits_{n = 1}^\infty {a_n } = s_n \therefore \sum\limits_{n = 1}^\infty {\frac{{a_1 + a_2 + \cdots + a_n }} {n}} = \sum\limits_{n = 1}^\infty {\frac{{s_n }} {n}} = s_n \sum\limits_{n = 1}^\infty {\frac{1} {n}}$

Therefore, $\displaystyle \sum\limits_{n = 1}^\infty {\frac{{a_1 + a_2 + \cdots + a_n }} {n}}$ is a divergent harmonic series?

What about this $\displaystyle \sum_{n=1}^{\infty}\frac{a_1+a_2+\cdots+a_n}{n}\si m\sum_{n=1}^{\infty}\frac{a_n}{n}\leqslant\sum_{n= 1}^{\infty}a_n$
• Nov 15th 2008, 01:30 PM
flyingsquirrel
Quote:

Originally Posted by Mathstud28
$\displaystyle \sum_{n=1}^{\infty}\frac{a_1+a_2+\cdots+a_n}{n}\si m\sum_{n=1}^{\infty}\frac{a_n}{n}$

What do you mean by $\displaystyle \sim$ ?
• Nov 15th 2008, 05:34 PM
Mathstud28
Quote:

Originally Posted by flyingsquirrel
What do you mean by $\displaystyle \sim$ ?

Sorry, I misread the problem. But here is how I would present the solution

Since this is a decreasing sequence of positive numbers let us apply Cauchy 'sCondensation test.

$\displaystyle \sum_{n=1}^{\infty}\sum_{k=1}^{n}a_k\frac{1}{n}$ converges iff $\displaystyle \sum_{n=1}^{n}2^n\cdot\sum_{k=1}^{2n}a_k\frac{1}{2 ^n}=\sum_{n=1}^{\infty}\sum_{k=1}^{2^n}a_k$ converges. From there it should be pretty obvious what the answer is.