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Proof by contradicion Sequences!

  1. Jan 8, 2009 #1
    Well, implicitly in a problem i came accros something that looks like it first requires to establish the following result:(to be more precise, the author uses the following result in the problem)

    Let [tex]\{y_n\}[/tex] be a sequence. If

    [tex]\lim_{n\rightarrow \infty}\frac{y_{n+1}}{y_n}=0,=> \lim_{n\rightarrow \infty}y_n=0[/tex]

    I tried a proof by contradicion, but i couln't really prove it.

    SO, any hints?
  2. jcsd
  3. Jan 8, 2009 #2
    Re: Sequences!

    Ok,here it is a thought, i'm not sure whether this would work though.

    Suppose that, [tex]lim_{n\rightarrow\infty}y_n=a[/tex] where a is different from zero.

    Then, we know that [tex]lim_{n\rightarrow\infty}y_{n+1}=a[/tex] as well.

    So, now we would have:


    Which is clearly a contradiction.

    Will this work?
  4. Jan 8, 2009 #3
    Re: Sequences!

    you still leave the possibility that y(n) doesn't converge at all.
  5. Jan 8, 2009 #4
    Re: Sequences!

    Well, that's not a problem. I forgot to mention, that i can show that y_n is a monotonic and bounded sequence, hence it must converge.
  6. Jan 8, 2009 #5
    Re: Sequences!

    right. yes, the proof is valid. (just to clarify the step, lim y(n+1) converges since y(n) converges and hence every subsequence must converge to the same limit)
  7. Jan 10, 2009 #6

    Gib Z

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    Homework Helper

    Re: Sequences!

    Just a thought, but if you are allowed to recognize the first limit as a positive result for the ratio convergence test, then the sum of the sequence must converge and hence the terms must approach zero as required.
  8. Jan 10, 2009 #7
    Re: Sequences!

    Well, that would work also in the case where y_n would be greater than zero for some n>N, and also in the case of alternating series, since in that case we could use the ratio test for alternating series:

    [tex]\lim_{n\rightarrow \infty}\left|\frac{y_{n+1}}{y_n}\right|=0<1[/tex] so the corresponding series

    [tex]\sum y_n[/tex] would converge absolutely and thus would be convergent as well, so y_n-->0 as n-->infinity.

    (p.s. i was basically aiming for a proof by contradiction in my first post)

  9. Jan 10, 2009 #8


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    Gold Member

    Re: Sequences!

    It's really rather simple when you think of it.
    let's go with definitions:

    lim (y_n/y_n-1)=L iff for every e>0 there exists N s.t for every n>N -e<y_n/y_n-1-L<e
    we assume that y_n doesn't vanish for each n.
    y_n=(y_n/y_n-1)*(y_n-1/y_n-2)*....*(y_2/y_1)*y_1 the ratio sequence converges to L thus it is bounded let M be this bound, assume that n-k=N for some k natural, then y_n-L=(y_n/y_n-1)*(y_n-1/y_n-2)*....*(y_2/y_1)*y_1-L

    y_n<=y_1*M^k(y_n/y_n-1)...(y_n-k+1/y_n-k)<y_1*M^k(L+e)^(n-k) as n->infinity y_n->0.

    Did I do a mess with this question, or am I on the right track, I had this in my first year at university so I should remember.
  10. Jan 10, 2009 #9
    Re: Sequences!

    Hmm..well, you are saying on this line that y_n is bounded by:


    but, if n-->infinity, how can we be sure that

    [tex](L+\epsilon)^{n-k}-->0[/tex] since obviously M^k doesn't go to zero.

    unless, [tex]|L+\epsilon|<1[/tex] of which is not necessarly true in general.

    I am probbably missing something in this whole picture, since i didn't analyze it a lot.
  11. Jan 10, 2009 #10


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    Gold Member

    Re: Sequences!

    I shoud relook my notes of first year if I havent thrown them away, I remember asking my TA this question, and it was also in courant and fritz text.
    There is some trick in it that I forgot, perhaps someone here remembers the proof that goes via epsilon.
  12. Jan 11, 2009 #11
    Re: Sequences!

    This is how I would tackle the problem.

    [tex]\lim_{n \rightarrow \infty} \frac{y_n_+_1}{y_n} = \lim_{n \rightarrow \infty} y_n_+_1 * \lim_{n \rightarrow \infty} \frac{1}{y_n} = 0 [/tex], implying that either of these limits converge to 0. If [tex] \lim_{n \rightarrow \infty} y_n_+_1 = 0 [/tex] then [tex] \lim_{n \rightarrow \infty} y_n = 0 [/tex] and we're done. So we take a look at the other limit. Suppose that [tex]\lim_{n \rightarrow \infty} \frac{1}{y_n} = 0 [/tex]. Then [tex] 1 = 0*\lim_{n \rightarrow \infty} y_n[/tex]. Since [tex]y_n[/tex] converges, say to "a", then we would have 1 = 0*a = 0, an obvious contradiction. Thus [tex]\lim_{n \rightarrow \infty} y_n_+_1 = 0[/tex] implying that [tex]\lim_{n \rightarrow \infty} y_n = 0 [/tex].
  13. Jan 11, 2009 #12
    Re: Sequences!

    I don't think you can separate the limits this way at first place, since if you are doing so, you are assuming that the limit of y_n as n goes to infinity exists, of which we are not sure, and we have to prove as well. So, i don't think this would work.
  14. Jan 11, 2009 #13
    Re: Sequences!

    I was thinking about that when I first thought of it, but then I saw you write this "I forgot to mention, that i can show that y_n is a monotonic and bounded sequence, hence it must converge. "
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