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anonymous

  • 4 years ago

please help!! How do you solve this sigma notation with an infinite series?

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  1. anonymous
    • 4 years ago
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    \[\sum_{n=1}^{\infty} 1/3^n\] ---please help!!

  2. anonymous
    • 4 years ago
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    This is the sum of a geometric sequence: \[\sum_{n=1}^\infty x^n = \frac{x}{1-x} \] so long as \[|x| < 1\] What is x in this case?

  3. anonymous
    • 4 years ago
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    that's the thing! in the question, there is only a fraction (1/3) and n being an exponent of the denominator. it's not like the usual notation. the answer is suppose to be 1/2 but i dont know why.

  4. anonymous
    • 4 years ago
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    \[\left(\frac{1}{3}\right)^n = \frac{1}{3^n} \]

  5. anonymous
    • 4 years ago
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    it is the case of convergence?jemurray

  6. anonymous
    • 4 years ago
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    does that mean that there is no x?

  7. anonymous
    • 4 years ago
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    It means x=1/3. What about convergence, tanusingh?

  8. anonymous
    • 4 years ago
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    like does this infinte series converge as n tends to infinitey

  9. anonymous
    • 4 years ago
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    \[\sum_{n=1}^{\infty}1/3^{n} = (1/3)/(1-(1/3))<1/2\]

  10. anonymous
    • 4 years ago
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    u can also use the formula a^n+1-1/a-1

  11. anonymous
    • 4 years ago
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    Yes it does, because |x| = 1/3 < 1

  12. anonymous
    • 4 years ago
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    amir that should be =, not <, right?

  13. anonymous
    • 4 years ago
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    like 1/3^n+1-1/1/3-1....then apply limit

  14. anonymous
    • 4 years ago
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    \[s _{n} = a/(1-q)\] a is the first term and q is \[(1/a ^{n+1})/(1/a ^{n})\]

  15. anonymous
    • 4 years ago
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    ya.....amir is right

  16. anonymous
    • 4 years ago
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    thanks! i understand it now!! my teacher didnt explain to us that the sum of the sigma notation equals \[x/1-x\]

  17. anonymous
    • 4 years ago
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    There's a lot flying around on this post.... for a geometric series, you have that \[\sum_{n=0}^\infty = \frac{1}{1-x} \] or, if you start from n = 1, \[\sum_{n=1}^\infty \ \frac{x}{1-x} \] In both cases this is valid iff |x| < 1.

  18. anonymous
    • 4 years ago
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    Its the sum of a infinite geometric series.

  19. anonymous
    • 4 years ago
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    was my ans wrong?

  20. anonymous
    • 4 years ago
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    tanusingh I'm not sure exactly what your answer was, could you clarify?

  21. anonymous
    • 4 years ago
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    i actually said that the sum=a^n+1-1/a-1

  22. anonymous
    • 4 years ago
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    Do you mean the nth partial sum?

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