Register to reply 
Prove or give a counter example is sum ai and sum bi are convergent series with nonn 
Share this thread: 
#1
Mar912, 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 nonnegative terms then sum aibi converges I believe that if it doesn't say "nonnegative terms" then this wouldn't be true. Am I correct? Since each of two nonnegative series converges then the series sum ai bi converges also. However I am not sure how to prove this. thanks 


#2
Mar912, 11:38 AM

Sci Advisor
HW Helper
PF Gold
P: 3,172




#3
Mar912, 11:48 AM

P: 25

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 


#4
Mar912, 11:56 AM

Sci Advisor
HW Helper
PF Gold
P: 3,172

Prove or give a counter example is sum ai and sum bi are convergent series with nonn



#5
Mar912, 01:30 PM

P: 25

the CauchySchwarz inequality:
sum ai bi is less then or equal to sum (ai^2)^1/2 sum (bi^2)^1/2 


#6
Mar912, 02:21 PM

Sci Advisor
HW Helper
PF Gold
P: 3,172

[tex]\sum a_i^2[/tex] if you know that [tex]\sum a_i[/tex] is finite? 


#7
Mar912, 02:58 PM

P: 25

if ai converges (finite) then (ai)2 converges also.



#8
Mar912, 03:00 PM

Sci Advisor
HW Helper
PF Gold
P: 3,172

And can you see how to use this fact along with the CauchySchwarz inequality to solve the problem? 


#9
Mar912, 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_(i1)+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 


#10
Mar912, 08:21 PM

Sci Advisor
HW Helper
PF Gold
P: 3,172

Let's look at the CauchySchwarz inequality again, which looks like the following assuming that [itex]a_i[/itex] and [itex]b_i[/itex] are nonnegative: [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]? 


#11
Mar912, 10:39 PM

Sci Advisor
HW Helper
Thanks
P: 25,246

Actually you don't really need CauchySchwarz 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.



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 nonnegative terms, a counterexample 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 PSeries (Cauchy sequences)  Calculus & Beyond Homework  1 