Compute each sum below. Give exact values, not decimal approximations. If the sum does not exist, click on "No sum".?

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  • geezer
    Lv 7
    1 month ago

    Why do people asking other people to do their homework for them NEVER say ''Please'' ?

    Thumb down !

    I've read your question again and NO .. I can't see the word PLEASE there !!

  • 1 month ago

    sum_(n=1)^∞ (4/9)^n = 4/5

  • 1 month ago

    Hope that you are expecting the sum to infinity in the series, (4/9)+(4/9)^2+...,

    Let give a little modification by adding 1 and substracting the same. 

    Then, -1 + 1+ (4/9)+(4/9)^2+...,

    1+ (4/9)+(4/9)^2+...,is a geometric series & the common ration, r = 4/9 <1, and a = 1

    Therefore sum to infinity = a/(1-r) = 1/[1-4/9] = 9/5.

    Thus the sum of the given series = 9/5 - 1= 4/5.

    Second series is also a geometric series if you add 1. Then r = -4 <1 and a = 1,

    As above sum to infinity = a/(1-r) = 1/(1+4) = 1/5

    Therefore sum to infinity of the given series = 1/5 - 1 = - 4/5.

  • 1 month ago

    Sum (4/9)^n = 4/5

    But, you should learn the technique, not ask us to do your homework

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  • 1 month ago

    The first one is (4/9)/(1 - 4/9) = 4/5.

    The second one diverges, i.e., "No sum"

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