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tux

  • 4 years ago

Decide convergence / divergence: sum n=1 to infinity 1/(n*sqrt(n)+2)

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  1. anonymous
    • 4 years ago
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    converges for sure

  2. anonymous
    • 4 years ago
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    since exponent in denominator is greater than 1

  3. tux
    • 4 years ago
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    How to solve using comparison test or integral test?

  4. amistre64
    • 4 years ago
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    integral test is just integrate from 1 to infinity

  5. amistre64
    • 4 years ago
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    comparison is a limit of the ratio of .... \[\frac{a_{n+1}}{a_n}\]

  6. anonymous
    • 4 years ago
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    actually that is "ratio test" comparison means compare to something else you know converges (or diverges)

  7. amistre64
    • 4 years ago
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    of is the comparison a b_n of another ..... yeah lol

  8. anonymous
    • 4 years ago
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    \[\sum_1^{\infty}\frac{1}{n\sqrt{n}+2}\]compare \[\frac{1}{n\sqrt{n}+2}<\frac{1}{n^{\frac{3}{2}}}\] second sum converges

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