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Math Help - Please help to determine whether the series is convergent or divergent

  1. #1
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    Please help to determine whether the series is convergent or divergent

    Please help to determine whether the series is convergent or divergent by Ratio Test
    2^(n+1)/2(n+2)^2+1
    When I found lim a(n+1)/a(n) = lim 2 (n is approaching infinity) =2
    2 is greater than 1, series converges.
    I probably made a mistake in lim calculations, if 2 is incorrect, will you please show me how to do it.
    Thank you very much
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  2. #2
    Member Abu-Khalil's Avatar
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    Is this \sum_{n}^\infty\frac{2^{n+1}}{2(n+2)^2+1}?

    \lim_{n\to\infty}\frac{a_{n}}{a_{n-1}}=\lim_{n\to\infty}\frac{2^{n+1}}{2(n+2)^2+1}\fr  ac{2(n+1)^2+1}{2^n}=2\lim_{n\to\infty}\frac{2n^2+4  n+3}{2n^2+8n+5}=2\cdot 1 = 2.

    So the serie diverges.

    In fact a_n\geq \frac{2^{n+1}}{2(n+2)^2+2}=\underbrace{\frac{2^n}{  n^2+4n+5}}_{n\to\infty,\to\infty}
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  3. #3
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    converges or diverges

    I am sorry, I typed the problem wrong, it is 2^n/2n^2 + 1. I am very sorry. Can you explain this one.
    Thank you
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  4. #4
    Member Abu-Khalil's Avatar
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    It's easier than it's looks like: If the serie converges \Rightarrow a_n\to 0 then if \neg(a_n\to 0)\Rightarrow the serie diverges.

    \lim_{n\to\infty}\frac{2^n}{2n^2+1}=\lim_{n\to\inf  ty}\frac{2^n\log 2}{4n}=\lim_{n\to\infty}\frac{2^n\log 2\log 2}{4}\to\infty
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  5. #5
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    Please solve it by ratio test

    I am sorry, but I do not understand it.

    Will you please show how to solve it by ratio test 2^n/2n^2 + 1.
    I did lim a(n+1) /a(n)= lim 2 =2
    The series diverges because 2 is greater than 1
    Is this correct?
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  6. #6
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    Yes, that's correct, but we can just bound the general term.

    For each n\ge1 it's \frac{2^{n}}{2n^{2}+1}>\frac{n}{2n^{2}+2n^{2}}=\fr  ac{1}{4}\cdot \frac{1}{n}, which diverges by direct comparison test with the harmonic series.
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  7. #7
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    Ratio Test

    Will you please check for correctness. Thank you a lot.

    Determine whether the series is convergent or divergent by Ratio Test.

    series (n+1)/n^2
    By Ratio Test lim is 1 , so the ratio test failed. We can use Limit Comparison. By Limit Comparison Test lim is 1, which is greater than 0, so both series are divergent based on b(n), because b(n) is divergent.

    Please explain if I am wrong. Thank you again
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  8. #8
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    That's correct, provided that b_n=\frac1n.

    You can also bound: \frac{n+1}{n^{2}}> \frac{n}{n^{2}}=\frac{1}{n}, and the conclusion follows.
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