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finding sum of infinite series using sigma notion

  • 26-03-2016 12:36PM
    #1
    iCrazzy
    Registered Users, Registered Users 2 Posts: 204 ✭✭


    Anyone could tell me how do I work with this type of question.
    Have only dealt with like variables e.g. r or n. but not 2 in one qs(n and x)


Comments

  • #2
    skippy1977
    Registered Users, Registered Users 2 Posts: 284 ✭✭skippy1977


    Because the term in bracket is less than 1 (the denominator is a greater number than the numerator) we can use the sum to infinity (of a geometric series) formula in the tables.
    a/(1-r)

    When given the sigma notation it is a good idea to write out the first few terms of the series so as to see what is going on...the sigma notation means the sum of all the series starting at n=0 as far as infinity...
    a is the first term....this series starts at n=0....so subbing that in gives a power of 0....so a=1
    Sub in n=1 to find the 2nd term which is 3x/(3x+4), the third term is (3x/(3x+4))^2...and so on...
    We can see so that r=3x/(3x+4)

    Put these two values into the sum to infinity formula and work it out...

    ...the answer is (3x+4)/4


  • #3
    iCrazzy
    Registered Users, Registered Users 2 Posts: 204 ✭✭iCrazzy


    skippy1977 wrote: »
    Because the term in bracket is less than 1 (the denominator is a greater number than the numerator) we can use the sum to infinity (of a geometric series) formula in the tables.
    a/(1-r)

    When given the sigma notation it is a good idea to write out the first few terms of the series so as to see what is going on...the sigma notation means the sum of all the series starting at n=0 as far as infinity...
    a is the first term....this series starts at n=0....so subbing that in gives a power of 0....so a=1
    Sub in n=1 to find the 2nd term which is 3x/(3x+4), the third term is (3x/(3x+4))^2...and so on...
    We can see so that r=3x/(3x+4)

    Put these two values into the sum to infinity formula and work it out...

    ...the answer is (3x+4)/4

    Ah I see, thank you.


  • #4
    iCrazzy
    Registered Users, Registered Users 2 Posts: 204 ✭✭iCrazzy


    skippy1977 wrote: »
    Because the term in bracket is less than 1 (the denominator is a greater number than the numerator) we can use the sum to infinity (of a geometric series) formula in the tables.
    a/(1-r)

    When given the sigma notation it is a good idea to write out the first few terms of the series so as to see what is going on...the sigma notation means the sum of all the series starting at n=0 as far as infinity...
    a is the first term....this series starts at n=0....so subbing that in gives a power of 0....so a=1
    Sub in n=1 to find the 2nd term which is 3x/(3x+4), the third term is (3x/(3x+4))^2...and so on...
    We can see so that r=3x/(3x+4)

    Put these two values into the sum to infinity formula and work it out...

    ...the answer is (3x+4)/4

    wait but should if a is 1 and r is 3x/3x+4
    then when you sub it in you should get 1/1-3x/3x+4


  • #5
    skippy1977
    Registered Users, Registered Users 2 Posts: 284 ✭✭skippy1977


    Yep and then tidy up...

    EDCO-E.png

    On the bottom find a common denominator...and then dividing by a fraction is the same as multiplying by that fraction turned upside down.


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