Register to reply

Prove or give a counter example is sum ai and sum bi are convergent series with non-n

by math25
Tags: convergent, counter, nonn, prove, series
Share this thread:
math25
#1
Mar9-12, 11:28 AM
P: 25
Hi

Can someone please help me to
prove or give a counter example is sum ai and sum bi are convergent series with non-negative terms then sum aibi converges


I believe that if it doesn't say "non-negative terms" then this wouldn't be true. Am I correct?

Since each of two non-negative series converges then the series sum ai bi converges also. However I am not sure how to prove this.

thanks
Phys.Org News Partner Science news on Phys.org
Mysterious source of ozone-depleting chemical baffles NASA
Water leads to chemical that gunks up biofuels production
How lizards regenerate their tails: Researchers discover genetic 'recipe'
jbunniii
#2
Mar9-12, 11:38 AM
Sci Advisor
HW Helper
PF Gold
jbunniii's Avatar
P: 3,288
Quote Quote by math25 View Post
Hi

Can someone please help me to
prove or give a counter example is sum ai and sum bi are convergent series with non-negative terms then sum aibi converges


I believe that if it doesn't say "non-negative terms" then this wouldn't be true. Am I correct?
Correct. Can you find a counterexample? (Hint: you can find one with [itex]a_i = b_i[/itex].)

Since each of two non-negative series converges then the series sum ai bi converges also. However I am not sure how to prove this.
Do you know any inequalities that involve [itex]\sum a_i b_i[/itex]?
math25
#3
Mar9-12, 11:48 AM
P: 25
Quote Quote by jbunniii View Post
Correct. Can you find a counterexample? (Hint: you can find one with [itex]a_i = b_i[/itex].)


ai= (cos n pie)/squareroot (n)= bi this would work right?


Do you know any inequalities that involve [itex]\sum a_i b_i[/itex]?

ai= (cos n pie)/squareroot (n)= bi this would work right?

I'm sorry I am not sure what you mean....i feel like this problem its very simple but for some reason I have such a hard time with it

jbunniii
#4
Mar9-12, 11:56 AM
Sci Advisor
HW Helper
PF Gold
jbunniii's Avatar
P: 3,288
Prove or give a counter example is sum ai and sum bi are convergent series with non-n

Quote Quote by math25 View Post
ai= (cos n pie)/squareroot (n)= bi this would work right?
Yes, that's the example I had in mind. Of course [itex]\cos(n\pi) = (-1)^n[/itex].

I'm sorry I am not sure what you mean....i feel like this problem its very simple but for some reason I have such a hard time with it
Do you know the Cauchy-Schwarz inequality?
math25
#5
Mar9-12, 01:30 PM
P: 25
the Cauchy-Schwarz inequality:


sum ai bi is less then or equal to sum (ai^2)^1/2 sum (bi^2)^1/2
jbunniii
#6
Mar9-12, 02:21 PM
Sci Advisor
HW Helper
PF Gold
jbunniii's Avatar
P: 3,288
Quote Quote by math25 View Post
the Cauchy-Schwarz inequality:


sum ai bi is less then or equal to sum (ai^2)^1/2 sum (bi^2)^1/2
OK, good. Now, given that all the [itex]a_i[/itex] are nonnegative, what can you say about

[tex]\sum a_i^2[/tex]

if you know that

[tex]\sum a_i[/tex]

is finite?
math25
#7
Mar9-12, 02:58 PM
P: 25
if ai converges (finite) then (ai)2 converges also.
jbunniii
#8
Mar9-12, 03:00 PM
Sci Advisor
HW Helper
PF Gold
jbunniii's Avatar
P: 3,288
Quote Quote by math25 View Post
if ai converges (finite) then (ai)2 converges also.
Correct, but do you know how to prove it?

And can you see how to use this fact along with the Cauchy-Schwarz inequality to solve the problem?
math25
#9
Mar9-12, 06:29 PM
P: 25
Since I am not sure how to use the Cauchy ineq. , how about something like this :

Let
Ai=Sum[i=0 to inf] ai
Bi=Sum[i= 0 to inf] bi
Ci=Sum[i=0 to inf] ai* bi
Ai, Bi, Ci are obviously strictly increasing (1) , because
Ai=A_(i-1)+a_n, similary Bi and Ci.
Let lim(n->inf)Ai=X, lim(n->inf)Bi=Y (because they converge).
Because they are strictly increasing,
=>Ai=X and Bi=Y for every i.
Ci=Ai*Bi, because
a1*b1+a2*b2+...an*bn<(a1+a2+..+an)*
*(b1+b2+..+bn)
From this, Ci=Ai*Bi=X*Y (2)
From (1) and (2) (monotonous and limited) Ci is convergent
jbunniii
#10
Mar9-12, 08:21 PM
Sci Advisor
HW Helper
PF Gold
jbunniii's Avatar
P: 3,288
Quote Quote by math25 View Post
Since I am not sure how to use the Cauchy ineq. , how about something like this :

Let
Ai=Sum[i=0 to inf] ai
Bi=Sum[i= 0 to inf] bi
Ci=Sum[i=0 to inf] ai* bi
These definitions don't make any sense. On the right hand side, you have summed over i = 0 to infinity. The result therefore does not depend on i.

Ai, Bi, Ci are obviously strictly increasing (1) ,
Since your definition of Ai, Bi, Ci above doesn't actually depend on i, this statement also makes no sense. A constant can't be strictly increasing.

Let's look at the Cauchy-Schwarz inequality again, which looks like the following assuming that [itex]a_i[/itex] and [itex]b_i[/itex] are non-negative:

[tex]\sum a_i b_i \leq \sqrt{\sum a_i^2} \sqrt{\sum b_i^2}[/tex]

Therefore, if [itex]\sum a_i^2[/itex] and [itex]\sum b_i^2[/itex] are finite, then so is [itex]\sum a_i b_i[/itex].

We know that [itex]\sum a_i[/itex] and [itex]\sum b_i[/itex] are finite. If you can show that this implies that [itex]\sum a_i^2[/itex] and [itex]\sum b_i^2[/itex] are finite, then you're done. So focus on this step.

Here's a hint: if [itex]x[/itex] is a nonnegative real number, what has to be true of [itex]x[/itex] in order to have [itex]x^2 \leq x[/itex]?
Dick
#11
Mar9-12, 10:39 PM
Sci Advisor
HW Helper
Thanks
P: 25,228
Actually you don't really need Cauchy-Schwarz do you? If the series ai converges then ai approaches 0. So ai<1 for i large enough, doesn't it? Use a comparison test.
jbunniii
#12
Mar10-12, 12:38 AM
Sci Advisor
HW Helper
PF Gold
jbunniii's Avatar
P: 3,288
Quote Quote by Dick View Post
Actually you don't really need Cauchy-Schwarz do you? If the series ai converges then ai approaches 0. So ai<1 for i large enough, doesn't it? Use a comparison test.
Yes, that's much better. When I see an inner product I always think "Cauchy-Schwarz" but in this case it meant that I didn't notice the more direct proof.


Register to reply

Related Discussions
If {S_n} is a sequence whose values lie inside [a,b], prove {S_n/n} is convergent. Calculus & Beyond Homework 4
How to prove a strictly diagonally dominant matrix is convergent Precalculus Mathematics Homework 1
Convergent series with non-negative terms, a counter-example with negative terms Calculus & Beyond Homework 1
Prove that a sequence of functions has a convergent subsequence Calculus & Beyond Homework 0
Divergent Harmonic Series, Convergent P-Series (Cauchy sequences) Calculus & Beyond Homework 1