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Analysis series proof

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data
    Let (a_n) and (b_n) be sequences with bn > 0 for all n in N. Assume that lim (an/bn) = L with L > 0.
    Prove that the series:
    [tex]\sum{a_n}[/tex] converges if and only if [tex]\sum{b_n}[/tex] converges


    2. Relevant equations



    3. The attempt at a solution

    Since this is an if and only if proof, i know that I need to first assume if the an series converges, then so does the bn series. Next, I need to show if I assume the second series converges, then so does the first.

    Alright, this is what I tried so far:

    Assume [tex]\sum{a_n}[/tex] converges.
    Meaning that the sequence (an) converges to zero.

    If (an) converges to zero that means that for epsilon greater than zero there exists an N in N so that for n>=N:

    |an|<epsilon

    Since we are assuming lim (an/bn) converges to L>0 that means for epsilon greater than zero, there exists an N in N so that for n>=N:
    |(an/bn)-L|<epsilon

    I think that to show this, I need to show that bn converges to zero as well, but I'm not too sure how to do that. Or also, there is a theorem stating if an and bn are sequences satisfying 0<=an<=bn for all n in N then
    if [tex]\sum{b_n}[/tex] converges then [tex]\sum{a_n}[/tex] converges.

    So I guess if I can show that an is larger than bn for all n in N then I can prove that the series converges as well.

    Any help/hints would be great!!!
     
  2. jcsd
  3. Oct 11, 2009 #2
    Just because bn converges to zero, that does not mean the series of bn will converge. It looks like the theorem you stated might be a good approach, but I'm not for sure. Note that if you can show bn≤can, for some constant c, that might help. Because if an converges, then can converges as well. I haven't worked this proof out, but these are just some thoughts.
     
  4. Oct 11, 2009 #3
    this was what I was thinking.... since the limit of an/bn>0 that means that an has to be >=bn for all n in N.
    We're also assuming bn>0 for all n in N which would mean
    0<=bn<=an for all n in N
    Which would mean if we assume that the series of an converges, that implies that the series of bn would converge.

    Is my assumption about the limits correct or am I way off?

    Also, then how would I go about proving that an converges if bn does if an is greater than or equal to bn for all n in N. Would I have to use the fact that if bn converges to zero if the series of bn converges?
     
  5. Oct 11, 2009 #4
    sorry i meant would i have to use the fact that if the series converges, then the sequence converges to zero
     
  6. Oct 11, 2009 #5
    Hi,

    I'll give you some hints to prove that if [itex]\sum{a_n}[/itex] converges, then so does [itex]\sum{b_n}[/itex]:

    1. When proving that an infinite sum converges, you can disregard any finite number of terms (the value of the sum might be affected, but not the convergence).
    2. If [itex]\sum{a_n}[/itex] converges and c is any constant, then [itex]\sum{ca_n}[/itex] also converges.
    3. If [itex]ca_n \geq 0[/itex] for some constant c and if [itex]|b_n| < ca_n[/itex] for all but finitely many n, then [itex]\sum{b_n}[/itex] converges (Comparison Test).
    4. We know that [itex]\lim_{x\rightarrow \infty}\frac{a_n}{b_n} = L > 0[/itex]. Use the [itex]\epsilon[/itex] definition of convergence with [itex]\epsilon = \frac{L}{2}[/itex]:

    There exists an N > 0 such that if n > N, then

    [tex]\left| \frac{a_n}{b_n} - L \right| < \frac{L}{2}[/tex]

    Simplify this inequality to conclude that

    [tex]0 < b_n < \frac{a_n}{\frac{L}{2}}[/tex]

    Now use observation #3 above to conclude that [itex]\sum{b_n}[/itex] converges. Note that the simplified inequality implies that [itex]a_n \geq 0[/itex] for all but finitely many [itex]a_n[/itex], justifying the use of #3.

    Let me know if this isn't clear, or if you would like more hints.

    Petek
     
  7. Oct 12, 2009 #6
    That's what confusing to me...
    How do you get from
    |an/bn-L|<L/2 to that inequality? I know that 0<bn, but how do yo justify that an/(L/2)>b?

    Also, to show that an converges if bn converges would I have to find a constant c so that
    0<c*seriesan< series of bn?
     
  8. Oct 12, 2009 #7
    or actually i meant 0<can<bn for all n in N
     
  9. Oct 12, 2009 #8
    oh wait nevermind, I understand, i was confused cause I was looking at the part showing that an<=(3/2)L*bn
    But that is what I will use to prove that if bn converges then so does an. Thanks so much for your help!
     
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