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Math Help - Limit of series

  1. #1
    Junior Member
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    Limit of series

    I'm just starting to study limits of sequences and have been asked to determine if the series;

    Sum (n=1 --> n = inf) 1/((n^2)+1) converges or diverges,

    Is it correct for me to do the following,

    Series will converge if the sequence (1/((n^2)+1) converges, hence;

    take lim (n-->inf) (1/((n^2)+1)

    and the limit = 0 as 1/inf is 0 in the limit.

    ?? is this correct? and should i be able to calculate the actual limit.

    Cheers.
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  2. #2
    Super Member
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    Quote Originally Posted by monster View Post
    I'm just starting to study limits of sequences and have been asked to determine if the series;

    Sum (n=1 --> n = inf) 1/((n^2)+1) converges or diverges,

    Is it correct for me to do the following,

    Series will converge if the sequence (1/((n^2)+1) converges, hence;

    take lim (n-->inf) (1/((n^2)+1)

    and the limit = 0 as 1/inf is 0 in the limit.

    ?? is this correct? and should i be able to calculate the actual limit.




    Cheers.
    Use comparsion test :  \frac{1}{n^2 + 1} < \frac{1}{n^2}

    BTW , the sum is  \frac{ \pi \coth(\pi) - 1}{2}
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  3. #3
    MHF Contributor

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    Quote Originally Posted by monster View Post
    I'm just starting to study limits of sequences and have been asked to determine if the series;

    Sum (n=1 --> n = inf) 1/((n^2)+1) converges or diverges,

    Is it correct for me to do the following,

    Series will converge if the sequence (1/((n^2)+1) converges, hence;
    You seem to be saying that the series converges if and only if the sequence converges but that not true. If the sequence a_n converges to any number other than 0, the series, \sum a_n does NOT converge. If the sequence converges to 0, the series may converge but not necessarily. for example 1/n converges to 0 but the series \sum 1/n does NOT converge.

    [/quote]take lim (n-->inf) (1/((n^2)+1)

    and the limit = 0 as 1/inf is 0 in the limit.

    ?? is this correct? and should i be able to calculate the actual limit.

    Cheers.[/QUOTE]
    No, that is not correct. As simple pendulum says, you can use the comparison test to show that this particular series does converge. You could also use the integral test: \int_0^\infty \frac{1}{x^2+ 1}= \left[arctan(x)]_0^\infty= \frac{\pi}{2} is finite so this series converges. There are, in general, no good ways to actually find the sum.
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