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Homework Help: True or False Analysis

  1. Jan 24, 2008 #1
    1. The problem statement, all variables and given/known data
    I need to prove if these are true, and provide a counter if they are false. Please tell me if I have these right, I think they are all true.

    (i) Let s(n) be a sequence s.t. [tex]\lim_{n\rightarrow\infty} (s(n+1) - s(n)) = 0[/tex]. Then s(n) must converge.

    (ii) Let s(n) be a sequence s.t. [tex]|s(n+1) - s(n)| < \frac{1}{n}[/tex] for all n. Then s(n) must converge.

    (iii) If [a(n)]^2 --> A^2, then either a(n) --> A or a(n) --> -A

    (iv) If [a(n)]^3 --> A^3, then a(n) --> A

    (v) If [s(n)]^2 --> S^2 and [tex]\lim_{n\rightarrow\infty} (s(n+1) - s(n)) = 0[/tex], then either s(n) --> S or s(n) --> -S

    2. Relevant equations

    3. The attempt at a solution
    (i) TRUE. It's a cauchy sequence. Proof involves showing that limit sup <= limit inf which implies limit sup = limit inf which in turn implies a limit exists.

    (ii) TRUE. A cauchy sequence with epsilon = 1/n

    (iii) TRUE. Trivial using the limit of the product of two convergent sequences is the product of the two limits.

    (iv) TRUE. Same way.

    (v) ??? Just look back at (i) and (iii), I guess...
    Last edited: Jan 24, 2008
  2. jcsd
  3. Jan 24, 2008 #2
    (i) is false take s(n) = sqrt(n), so s(n + 1) - s(n) = sqrt(n + 1)-sqrt(n) = 1/(sqrt(n + 1) + sqrt(n)) -> 0 as n->inf but clearly s(n) diverges
  4. Jan 24, 2008 #3
    Good catch, sqrt(n) is not cauchy. Thanks.
  5. Jan 24, 2008 #4
    also your reasoning in (iii) and (iv) is wrong, i think they are true though(I didn't verify)
  6. Jan 24, 2008 #5
    How is the reasoning for (iii) wrong? I was gonna argue this way.

    Assume lim = B != A and B != -A.
    the lim a^2 = lim a * lim a = B^2 = A^2.

    This implies B = +-A, which contradicts assumption.

    Where did I go wrong?
  7. Jan 24, 2008 #6
    I fixed (ii) by the way, there was a big typo...
  8. Jan 24, 2008 #7
    Sorry about that! (iii) is false, and your reasoning is wrong, we don't know a_n converges,
    try a_n = (-1)^n
  9. Jan 24, 2008 #8
    Oh, you are correct. I assumed a(n) converged.

    Likewise, I believe (ii) is wrong and I need to compare to the harmonic series, I"m still wondering as to how though.
  10. Jan 24, 2008 #9


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    Most of these can be dealt with by looking at the sequence (-1)^n.
  11. Jan 24, 2008 #10
    yes that works,

    note (ii) is true if you replace 1/n by 1/k^n for some constant k >1, or if you replace 1/n by a constant c with 0<c< 1.
  12. Jan 24, 2008 #11
    For (ii), i can use s(n) = ln(x) to prove that it's false, correct?
  13. Jan 24, 2008 #12
    very nice yes that seems to work using the fact ln is increasing and ln1 = 0

    and since you figured it out, here's how to do it with the harmonic series since you asked earlier,

    set a_n = sum(1/k, k = 1,..., n), then a_(n + 1) - a_n = 1/(n + 1) < 1/n, and lim n->inf a_n diverges
    Last edited: Jan 24, 2008
  14. Jan 24, 2008 #13
    (iv) is tricky. I tried using the difference of squares, that is

    [a(n)]^3 - A^3 = [a(n) - A]*[a(n)^2 + a(n)A + A^2]

    I feel like from there I can argue that the second bracket is bounded and positive, but I'm not sure that's valid.
  15. Jan 25, 2008 #14
    *sigh*, I have no clue how to prove (iv) and (v).

    I know they are true. Basically for (v) both restrictions tell me that the limiting behavior of the sequences are the same for consecutive terms (which eliminates the possibility of an oscilating sequence) and that it must be bounded (since the product of limits is convergent).
    Likewise, I know for (iv) that a(n) must be bounded, and I know it doesn't oscillate, although I'm not sure I can come up with a great rigourous explanation as to why (I can say that a(n) and a(n)^3 always have the same sign, so if a(n) oscilates in the limit the actual magnitudes of its values will change, thus so will a(n)^3).
  16. Jan 25, 2008 #15
    For iv, if you know that the cube root is a continuous function, you're done, I believe.

    And for v, maybe you could do something similar with the square root function, but I'm not sure.
  17. Jan 26, 2008 #16
    ok i'll give you a hint for (iv),
    the way to solve this is to notice |a_n - a| = |a_n^3 - a^3|/|a_n^2 + a_na + a^2| < |a_n^3 - a^3|/|a_na| if a_na > 0, but if a is positive then a_n is also positive for all n greater than some N, so a_na > 0 for all n > N, similiarly if a is negative a_n is also negative for all n greater than some N'.

    if a = 0 the result is trivial.

    if a > 0 , since a_n^3 -> a^3, there is N_1 s.t. for all n > N_1, |a_n^3 - a^3| < (a^3)/2, so a_n^3 > (a^3)/2 > 0, so a_n > a/2^(1/3), so 1/a_n < 2^(1/3)/a. Now notice for n > N_1, |a_n - a| = |a_n^3 - a^3|/|a_n^2 + a_na + a^2| < |a_n^3 - a^3|/|a_na| < ...
    (use convergence again to get another N_2 for a particular choice of e, take the max, call it N, continue the above)

    if a < 0, same trick.
    Last edited: Jan 26, 2008
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