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Math Help - Convergent or Divergent

  1. #1
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    Convergent or Divergent

    Right now we're doing Convergent and Divergent tests. We haven't gotten to Infinite Series' yet, but I know it's coming. I want to see if I have the right answer and make sure I have the right concepts.

    <br />
\int_{0}^{\infty} \frac {dx}{\sqrt{2x+1}}<br />

    I swap out infinite for t and integrate the equation using substitution and get

    <br />
\sqrt{2x+1}<br />

    Then I have

    <br />
\sqrt{2t+1}-\sqrt{2(0)+1}<br />

    Which is

    <br />
\sqrt{2t+1}-\sqrt{1}<br />

    Is this convergent? Because no matter what value of t, I will always be left with a finite number?
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  2. #2
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    \lim _{t \to \infty } \left( {\sqrt {2t + 1}  - 1} \right) = ?
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  3. #3
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    Sorry, I should have specified limits in the final answer.

    Quote Originally Posted by Plato View Post
    \lim _{t \to \infty } \left( {\sqrt {2t + 1}  - 1} \right) = ?
    \lim _{t \to \infty } \left( {\sqrt {2t + 1}  - 1} \right) = \infty? Which would make it Divergent? That seems too easy.
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  4. #4
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    Quote Originally Posted by maxreality View Post
    Right now we're doing Convergent and Divergent tests.
    i'd use at least two tests in order to find convergence or divergence without computing the integral, tell me if you've covered convergence tests.
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  5. #5
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    Quote Originally Posted by Krizalid View Post
    i'd use at least two tests in order to find convergence or divergence without computing the integral, tell me if you've covered convergence tests.
    We have, and I believe that's what I'm struggling with. I'm watching some of Patrick's tutorials on his YouTube channel trying to make sense of it all.
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