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mathew0135

  • 3 years ago

Check whether the series defined below is convergent and if so, find its sum.

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  1. mathew0135
    • 3 years ago
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    \[\sum_{n=1}^{\infty} \frac{ 2^n }{ (-12)^{n-1} }\] and the equation.

  2. zordoloom
    • 3 years ago
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    What math class is this for?

  3. etemplin
    • 3 years ago
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    likely calc 2, because i just had a test on this

  4. zordoloom
    • 3 years ago
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    BTW, this series converges.

  5. mathew0135
    • 3 years ago
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    Math 1B in Australia (presumably similar to calc 2 from what i've seen).

  6. zordoloom
    • 3 years ago
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    ok, the sum will also be -12/5

  7. mathew0135
    • 3 years ago
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    thanks, good to know but how do you work that out?

  8. zordoloom
    • 3 years ago
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    Take the limit as n goes to infinity.

  9. mathew0135
    • 3 years ago
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    thus far i've found the absolute value, then tried to find the limit as it approaches infinity. That's what i believe you do.

  10. zordoloom
    • 3 years ago
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    Ya. That would be correct. So, the form is an+1/an

  11. zordoloom
    • 3 years ago
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    And you plug in the formula that you have into this equation.

  12. zordoloom
    • 3 years ago
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    -(2^n+1)/(12^(n+1)(-1)) all over (2^n)/(-12^(n-1))

  13. mathew0135
    • 3 years ago
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    okay so because the is to a higher power the limit is equal to 0 and thus converges. Hate to ask but how do you find the sum. Think you tried to explain it there but i'm still a bit confused. does: (-(2^n+1)/(12^(n+1)(-1)))/((2^n)/(-12^(n-1))) find me the sum when n=1?

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