What is the Bound for ∑an2 and Why?

[saddlepoint]

Homework Statement

Show if a_n > 0 for each n\inN and if ∑a_n converges, then ∑a_n² converges and that ∑1/a_n diverges.

NB. all ∑ are between n=1 and ∞

Homework Equations

The Attempt at a Solution

Let partial sums of ∑a_n² be S_k = a₁ + ... + a_k

To say ∑a_n² is absolutely convergent is to say ∑|a_n²| is convergent.

It follows that partial sums T_k = |a₁²| + |a₂²| + ... + |a_k²| of the series are bounded above by M.

Then by an extended form of the Triangle Inequality we have:

|S_k|

= a₁² + a₂² + ... + a_k²

≤ |a₁²| + |a₂²| + ... + |a_k²|

= T_k

≤ M

Hence the sequence {|S_k²} is bounded above by M. It is bounded below by 0 as a_n > 0 so a_n² > 0. Therefore it is bounded.

It is therefore convergent.

Is this correct so far? Would a similar proof follow for showing ∑1/a_n is divergent?

Can anyone help with this please?

 

[PeroK]

I think you're missing a simple solution. Think of what happens to a_n as n gets large.

I don't follow your proof. (How are you using the convergence of the series a_n?)

 

[micromass]

Homework Statement

Show if a_n > 0 for each n\inN and if ∑a_n converges, then ∑a_n² converges and that ∑1/a_n diverges.

NB. all ∑ are between n=1 and ∞

Homework Equations

The Attempt at a Solution

Let partial sums of ∑a_n² be S_k = a₁ + ... + a_k

To say ∑a_n² is absolutely convergent is to say ∑|a_n²| is convergent.

It follows that partial sums T_k = |a₁²| + |a₂²| + ... + |a_k²| of the series are bounded above by M.

What is $M$ and why does it bound the $T_k$?
You have only established that $\sum a_n^2$ being absolutely convergent is the same as saying that $\sum|a_n|^2$ converges. But you have not established the fact that $\sum a_n^2$ actually absolutely converges!