How do I show the series of an*bn converges?

[LHS]

Homework Statement

[PLAIN]http://img11.imageshack.us/img11/4384/unledqr.png

Homework Equations

The Attempt at a Solution

How do I show the series of an*bn converges?

 

[hunt_mat]

Perhaps consider the partial sums of the series and treat that like a sequence.

 

[LHS]

how about

0<=(a-b)^2
2ab<=a^2+b^2
2ab<=a^2+b^2
therefore

2*sum(ab)<=sum(a^2)+sum(b^2)

Where the RHS converges

As the LHS converges absolutely sum(ab) converges?

Any ideas about the examples?
[PLAIN]http://img35.imageshack.us/img35/9916/unledis.png

this is from a past paper by the way, I can link you if you like, it's not a take home test

 

[jbunniii]

You have the right idea, but you should fill in a few details.

For example, you showed that

2ab <= a^2 + b^2

but you implicitly use the fact that

2|ab| <= a^2 + b^2

if you're talking about absolute convergence.

Why is the inequality true for |ab|? And do you need this? The statement that you are trying to prove doesn't have an absolute value on the left side.

Also, I recommend that you start by taking finite sums of both sides of the inequalities (summing from 1 to N), and then be very careful to explain why it remains true as N goes to infinity.

 

[LHS]

thank you very much for your time, I will do that :)

 

[jbunniii]

For the counterexample in the complex case, try to think of sequences such that

a_r^2
and
b_r^2

are both real with alternating signs, but

a_r b_r

is always real and positive.

 

[LHS]

So say something like an=i^n/n^0.5 and bn=an=-i^n/n^0.5

Excellent! that seems to work, thank you very much, now I can finally go to sleep

 

[jbunniii]

Yep, that's the example I was thinking of.

 

[Ray Vickson]

You can try to show that S(n) = sum{a_k*b_k,k=1..n} is a Cauchy sequence, that is, that S(m+n) - S(n) goes to zero for large n, m. Note that S(m+n)-S(n) is just the inner product of two vectors (a_n, a_{n+1},...,a_{n+m}) and (b_n,b_{n+1},...,b_{n+m}) in m-dimensional space.

RGV