Rearranging Series

  1. As we know the sum of the series ((-1)^(n-1))/(n) is equal to ln2: I have bin asked to rearrange the series so its sum is SQRT2



    Also write out the first 20 terms



    How should I go about solving. Set original series equal to SQRT 2 and solve from there? Use trial and error until I arrive at the correct series.
     
  2. jcsd
  3. micromass

    micromass 18,679
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    I'll give a small sketch. I'll leave the details to you.

    First take some positive terms such that the sum of these terms is just above sqrt(2).
    Then take some negative terms such that the sum of all the terms comes back below sqrt(2).
    Then take enough positive terms such that the sum comes above sqrt(2) again.
    Continue with this process...
     
  4. Thanks micromass Im actually gonna come back to that question. While I have u here I have bin posed the following:

    Describe the following series and if possible find its limits:

    1/4 +1/10+1/18+1/28.....

    any ideas??
     
  5. micromass

    micromass 18,679
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    Well, let's take a look at the sequence 4,10,18,28,...
    What is the fifth element of this sequence?
    Can you find a general rule for how to obtain the next element in this sequence?
    And: more important, can you find a general term for this sequence?
     
  6. Well from observing the sequence the next term is 1/40. Correct??
     
  7. micromass

    micromass 18,679
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    Yes, so the denumerators are the following:

    [tex]4,4+6,4+6+8,4+6+8+10,4+6+8+10+12,...[/tex]

    Now try to find a general rule for this sequence.

    HINT: you've probably seen that

    [tex]\sum_{k=0}^n{k}=\frac{n(n+1)}{2}[/tex]

    use this...
     
  8. Well as a rule for the sequence isnt it similar to fibonacci. SO:

    fn = fn-1+fn-2

    for n>1
     
  9. micromass

    micromass 18,679
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    No, this isn't like fibonacci at all. What is the general rule for this sequence?
     
  10. the next term is just the previous terms added to the new term.
     
  11. micromass

    micromass 18,679
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    Indeed. So, if you're given the nth term [tex]x_n[/tex]. How do you obtain the next term. In other words, fill in the blanks:

    [tex]x_{n+1}=x_n+...[/tex]

    HINT:
    x_0=4
    x_1=x_0+6
    x_2=x_1+8
    x_3=x_2+10
     
  12. I think I have the general rule is it:

    1/(n+2)^2 +2

    so sub in 0

    1/(0+2)^2 +0

    1/4

    sub in 1

    1/(1+2)^2 + 1

    1/10....
     
  13. micromass

    micromass 18,679
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    Yes, that is the general rule. So now you have the series

    [tex]\sum_{n=0}^{+\infty}{\frac{1}{(n+2)^2+n}}[/tex]
     
  14. Great it says find its limit? Is this the sum?
     
  15. micromass

    micromass 18,679
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    Well, to find the limit of a series is not always possible. But in this case it is.
    Try to write the series as a telescopic sum. To do that, split up the fraction

    [tex]\frac{1}{(n+2)^2+n}[/tex]

    in partial fractions. This will help a great deal...
     
  16. Oh ok should it look like:

    1 = A/(n+2)^2 +B/(n+2) + C/n
    Then find values for A B and C?
     
  17. Giving values of

    A = -1/4 B=/1/2 c=1/4
     
  18. micromass

    micromass 18,679
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    Nonono. First you need to simplify [tex](n+2)^2+n[/tex]. Then you need to factor it. Only then can you begin calculating the partial fractions...
     
  19. Ah ok so that gives n^2 +5n +4 which factors to:

    (n+1)(n+4)

    Then solving will give values for A and B. Then wot shall i do???
     
  20. micromass

    micromass 18,679
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    Yes, this is correct. So what are your values for A and B?
    What is

    [tex]\frac{1}{(n+1)(n+4)}=...[/tex]?
     
  21. A = 1/3
    b= -1/3
     
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