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anonymous

  • 5 years ago

determine whether this series converges or diverges sigma 1 to infinity -1^(n-1)2^n/(n^2)

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  1. anonymous
    • 5 years ago
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    These fractional types usually suggest ratio test

  2. anonymous
    • 5 years ago
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    \[\sum_{n=1}^{\infty}\frac{(-1)^{n-1} 2^n}{n^2}\] as chag said, using the ratio test, you'll get: \[|\frac{an+1}{an}| = |\frac{(-1)^n2^{n+1}}{(n+1)^2} . \frac{n^2}{(-1)^{n-1}2^n}|\]

  3. anonymous
    • 5 years ago
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    so :

  4. anonymous
    • 5 years ago
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    \[= \frac{2^n.2^1}{(n+1)^2} . \frac{n^2}{2^n} = \frac{2n^2}{(n+1)^2}\] now the second step is to find the limit as n--> infinity: \[\lim_{n \rightarrow \infty} \frac{2n^2}{(n+1)^2} = 2\] and the Ratio Test's theorem says the following: if L (limit) > 1 = series diverge if L <1 = series converge if L = 1 , no conclusion Correct me if I'm wrong please ^_^

  5. anonymous
    • 5 years ago
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    so for this case, since 2 > 1, then the following series diverge :)

  6. anonymous
    • 5 years ago
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    if you don't understand what I did, let me know :)

  7. anonymous
    • 5 years ago
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    ty i got it

  8. anonymous
    • 5 years ago
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    np ^_^

  9. anonymous
    • 5 years ago
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    can u help me with another problem

  10. anonymous
    • 5 years ago
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    I'll give it a try ^_^

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spraguer (Moderator)
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