1. Another Convergence Problem

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

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

My Attempt:

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

2. Hi,
Originally Posted by RedBarchetta
$\sum\limits_{n = 1}^\infty {\frac{{s_n }}
{n}} = s_n \sum\limits_{n = 1}^\infty {\frac{1}
{n}}
$
You can't do this, $s_n$ depends on $n$ !

Hint : for all $n\geq 1$, $s_n\geq a_1>0$ because $a_k>0$ for all $k\geq 1$.

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

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

My Attempt:

$
\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, $
\sum\limits_{n = 1}^\infty {\frac{{a_1 + a_2 + \cdots + a_n }}
{n}}
$
is a divergent harmonic series?
What about this $\sum_{n=1}^{\infty}\frac{a_1+a_2+\cdots+a_n}{n}\si m\sum_{n=1}^{\infty}\frac{a_n}{n}\leqslant\sum_{n= 1}^{\infty}a_n$

4. Originally Posted by Mathstud28
$\sum_{n=1}^{\infty}\frac{a_1+a_2+\cdots+a_n}{n}\si m\sum_{n=1}^{\infty}\frac{a_n}{n}$
What do you mean by $\sim$ ?

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

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