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anonymous

  • 5 years ago

What series test do you use when the sum starts at n =1 (2 + (-1)^n)/(n(n^(1/2)))?

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  1. anonymous
    • 5 years ago
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    In the numerator you'll end up with 2+1, 2-1, 2+1 etc. but in the denominator you have n^(3/2) which will go to infinity as n gets large. So the terms will go to zero.

  2. anonymous
    • 5 years ago
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    so you use the p series test?

  3. anonymous
    • 5 years ago
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    well its like two different types put together - the denominator is a p-series

  4. anonymous
    • 5 years ago
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    thanks!

  5. anonymous
    • 5 years ago
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    what series test do you use for \[\sum_{n=1}^{\infty} (n-1)/(n4^n)\]

  6. anonymous
    • 5 years ago
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    You can break this one up into two parts...\[(n/n4^{n})-(1/n4^{n})\] In both series, the terms will approach zero.

  7. anonymous
    • 5 years ago
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    Alright so you can break it up, but how do you tell if it converges or diverges using the various series tests?

  8. anonymous
    • 5 years ago
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    The first part simplifies to a geometric series\[(1/4)^{n}\]

  9. anonymous
    • 5 years ago
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    How did you get that to reduce to \[1/4^{n}\]

  10. anonymous
    • 5 years ago
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    because the n's cancel

  11. anonymous
    • 5 years ago
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    but the top is subtraction not multiplication? so you cant just cancel like that...

  12. anonymous
    • 5 years ago
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    I broke up the fraction into two parts because you have (n-1) in the numerator.

  13. anonymous
    • 5 years ago
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    \[(n-1)/(n4^{n}) = (n/n4^{n}) - (1/n4^{n})\]

  14. anonymous
    • 5 years ago
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    i still dont understand what you all cancel out to get the (1/4)^n?

  15. anonymous
    • 5 years ago
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    The first part (n/n4^n) - the n's will cancel and you're left with (1/4^n) = (1/4)^n - which is a geometric series.

  16. anonymous
    • 5 years ago
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    so you ignore the (1/n4^n)?

  17. anonymous
    • 5 years ago
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    You can't ignore it, but you can treat it as a separate series.

  18. anonymous
    • 5 years ago
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    ok so if im trying to find if the series is convergent or divergent... i represent with evidence using the geometric series as a whole? Because up til this section we are limited to the geometric, telescoping, limit comparison, direct comparison, p series and integral tests?

  19. anonymous
    • 5 years ago
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    I guess since you have a fraction you could try limit comparison.

  20. anonymous
    • 5 years ago
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    alright thanks

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