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Prove ##(1/(a_1+1) ) + (1/(a_2+1) ) ....##

  1. Jul 16, 2016 #1
    1. The problem statement, all variables and given/known data
    a1=3 an+1=0.5(an2+1)
    "n" is natural number

    prove that (1/(a1+1) ) + (1/(a2+1) ) ......... + (1/(an+1) ) <1/2
    is true for every n

    3. The attempt at a solution
    could not figure out anything
     
  2. jcsd
  3. Jul 16, 2016 #2

    haruspex

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    I would look for a simpler sequence, geometric say, which serves as a lower bound for the an for all n > some threshold N. Then it would be a matter of manually summing the 1/(an+1) up to N and adding the sum corresponding to the infinite bounding sequence after N.
     
  4. Jul 16, 2016 #3
    Not sure, maybe one trick would be to work with the sequence ##b_{n+1}=0.5b_n, b_1=3## or the sequence ##b_{n+1}=0.5b_n^2##. If you can prove that the corresponding sum for ##b_n## is <1/2 then the corresponding sum for ##a_n## also satisfies <1/2 because it is ##b_n+1<a_n+1 \rightarrow \frac{1}{a_n+1}<\frac{1}{b_n+1}##
     
  5. Jul 17, 2016 #4
    Ok my initial suggestion about ##b_n## don't work, try ##b_{n+1}=3b_n, b_1=3##. You can prove that ##b_n<a_n , n\geq 4##, and also you prove easily that ##\sum\limits_{n=1}^{\infty}\frac{1}{b_n}=\frac{1}{2}## and then using inequality ##\frac{1}{a_n+1}<\frac{1}{b_n+1}<\frac{1}{b_n}## for n>=4.
     
  6. Jul 17, 2016 #5

    haruspex

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    I feel my hint in post #2 was enough to be going on with. Please let the student have time to work with one hint before laying it out further.
     
  7. Jul 17, 2016 #6
    i could not find any rule for adding every number from N that i took as a threshold it is not geometric progression to add up every number and find limit of function
     
  8. Jul 17, 2016 #7
    thanks i guess it is right but may it be proof for sum of all 1/3^n this
    -1 + 1 + 1/3 + 1/3^2 + 1/3^n =
    -1 +(3^n+1 -1)/ 3-1) = 1-(1/3)^n/2
     
  9. Jul 17, 2016 #8
    and what about this
    2an+1 - 2 = an2- 1
    1/an+1= 1/an-1 + 1 / an+1
     
  10. Jul 17, 2016 #9

    haruspex

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    I did not say it is. I said try to find some other sequence bn, perhaps a geometric progression, and some number N, such that:
    • For n>N, bn>=an
    • Σ1N1/(an+1) + ΣN+11/(bn+1) < 1/2
     
  11. Jul 17, 2016 #10
    ok got it so the way that delta^2 offered
     
  12. Jul 17, 2016 #11

    haruspex

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    Yes. But I have not checked whether Delta^2's specific sequence and starting point work, or if they do, whether they're the simplest choices.
     
  13. Jul 17, 2016 #12
    the second post of him is right and easy also
     
  14. Jul 17, 2016 #13

    Ray Vickson

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    Bounding will need to be done VERY carefully, because the "1/2" limit is pretty strict.

    If we use the exact recursion ##a_1 = 3, a_{n+1} = \frac{1}{2} (a_n^2 + 1), n = 1,2, \ldots## and compute exact values up to about ##n = 12##, we see that the sequnece ##a_n## grows rapidly. Using Maple and exact arithmetic whenever it is practical, we get
    $$\begin{array}{l}
    a_{12} =
    150069266781857262292843669672858603\\
    4769448280147464726353368783536748476\\
    47440046640744854967963947245498612021\\
    477388413904375180451656270390552278692\\
    1452055548775397792172060973375592551323900\\
    39967096883676820619379899904933201414659759\\
    5705864255753847966680994896463920097666188\\
    3908504636929575700712046785222999071294587426\\
    7583554677334026682716307498157604046253018642062\\
    0448794191288742297786522568403222907220885\\
    \doteq .1500692668 \times 10^{418}
    \end{array} $$
    More to the point are the values of
    $$S_n = \sum_{j=1}^n \frac{1}{a_j+1} $$
    for ##n = 1,2, \ldots##.

    To 833 digit accuracy we find ##S_{12} = 0.5## (that is, 0.5 followed by 832 zeros), but to 834 digits of precision we find that ##S_{12}## is a tiny bit less than 0.5; it prints as 0.499..., which is 0.4 followed by 833 '9's . Of course, values of ##S_n## for ##n > 12## will be a tiny bit larger, and so the bound of 1/2---if true--- will be very, very close.

    Note added in edit: If
    $$ S_n = \frac{M_n}{N_n} $$
    for integers ##M_n, N_n## we find that ##N_n = 2 M_n+2##, at least for ##n## from 1 to 12. Thus,
    $$ S_n = \frac{1}{2} \frac{M_n}{M_n+1} $$
    for ##n = 1, 2, \ldots, 12##.

    The ##M_n## grow rapidly with ##n##; for example, ##M_{12}## is an 834-digit integer: ##M_{12} \doteq 5.630196208 \times 10^{833}##. Thus
    $$ S_{12} \doteq \frac{1}{2} \left( 1 - 1.776137035 \times 10^{-834} \right) . $$
     
    Last edited: Jul 17, 2016
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