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tux

  • 4 years ago

Decide convergence/divergence from n=1 to infinity (2^n+1)/(3^n-1)

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  1. anonymous
    • 4 years ago
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    converges for sure, and in fact i think the sum is not even 5

  2. anonymous
    • 4 years ago
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    ignore the irrelavent -1 and +1 and use the root test, get it right away

  3. tux
    • 4 years ago
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    It is hard to calculate using root test

  4. anonymous
    • 4 years ago
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    not when the index is in the exponent, then it is easy

  5. anonymous
    • 4 years ago
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    \[\sqrt[n]{2^n}=2\]!

  6. tux
    • 4 years ago
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    What to do with +1 and -1?

  7. tux
    • 4 years ago
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    Still have no idea how to do

  8. Zarkon
    • 4 years ago
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    if you want...you can use the comparison test

  9. tux
    • 4 years ago
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    Compare with (2/3)^n?

  10. Zarkon
    • 4 years ago
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    sure...use the limit comparison test then.

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