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frx

  • 3 years ago

Is the serie \[\sum_{n=0}^{\infty} (-1)^{n}\frac{ n+2 }{ 3^{n} }\] convergent or divergent? Calculate is't value. \[\lim_{n \rightarrow \infty} \left| \left( \frac{ (-1)^{n+1}(n+3) }{ 3^{n+1} } \right)\left( \frac{ 3^{n} }{ (-1)^{n}(n+2) } \right) \right|\] \[\frac{ 1 }{ 3 } \lim_{n \rightarrow \infty} \left| \frac{ n+3 }{n+2 } \right|=\frac{ 1 }{ 3 }\] \[\frac{ 1 }{ 3 } < 1; Convergent\] But how do I calculate it's value?

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  1. frx
    • 3 years ago
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    @phi

  2. phi
    • 3 years ago
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    I would try summing the negative terms and the positive terms

  3. frx
    • 3 years ago
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    Don't really know what you mean, should I expand the serie a couple of times then sum?

  4. Zarkon
    • 3 years ago
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    \[\sum_{n=0}^{\infty} (-1)^{n}\frac{ n+2 }{ 3^{n} }\] \[\sum_{n=0}^{\infty} n\left(\frac{-1}{ 3 }\right)^n+2\sum_{n=0}^{\infty} \left(\frac{-1}{ 3 }\right)^n\]

  5. Zarkon
    • 3 years ago
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    \[2\sum_{n=0}^{\infty} \left(\frac{-1}{ 3 }\right)^n\] is just a geometric sum...should be easy

  6. Zarkon
    • 3 years ago
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    for \[\sum_{n=0}^{\infty} n\left(\frac{-1}{ 3 }\right)^n\] write as \[\frac{-1}{3}\sum_{n=0}^{\infty} n\left(\frac{-1}{ 3 }\right)^{n-1}\] then as \[\frac{-1}{3}\sum_{n=0}^{\infty} n\left(x\right)^{n-1}\] integrate this wrt x then compute the sum... then differentiate and plug in -1/3

  7. Zarkon
    • 3 years ago
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    I get \[\frac{-3}{16}+\frac{3}{2}=\frac{21}{16}\]

  8. frx
    • 3 years ago
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    Your answer is correct according to my key, what's the metod you're using named? I think I need to study the concept a bit closer

  9. Zarkon
    • 3 years ago
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    if a series converges uniformly then it can integrated or differentiated term by term

  10. frx
    • 3 years ago
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    I think I've found the section in my book now, don't get the concept yet but now I know where to start working on it. Thank you!

  11. Zarkon
    • 3 years ago
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    good

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