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Math Help - limit of a sequence

  1. #1
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    limit of a sequence

    Hey people.
    I have a sequence {b_n} which is defined like that :
    b_n = 1+(1/3)+...+(1/1(2n-1)) , i.e b_n = sum (1/(2k-1)) from k=1 to n

    I need to prove that lim (b_n) as n->inf = inf
    I think I should use the squeeze theorem , but this sequence approaches infinity really slow, so I couldn't find a sequence {a_n} which approaches infinity and b_n >= a_n for almost every n...

    Any clues anyone?
    Thanks!
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  2. #2
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    Quote Originally Posted by Gok2 View Post
    Hey people.
    I have a sequence {b_n} which is defined like that :
    b_n = 1+(1/3)+...+(1/1(2n-1)) , i.e b_n = sum (1/(2k-1)) from k=1 to n

    I need to prove that lim (b_n) as n->inf = inf
    I think I should use the squeeze theorem , but this sequence approaches infinity really slow, so I couldn't find a sequence {a_n} which approaches infinity and b_n >= a_n for almost every n...

    Any clues anyone?
    Thanks!
    What you have posted is not a sequence, but a series.

    Are you trying to evaluate the n^{\textrm{th}} term of the series, or are you trying to show that the series is divergent?
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  3. #3
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    Quote Originally Posted by Gok2 View Post
    Hey people.
    I have a sequence {b_n} which is defined like that :
    b_n = 1+(1/3)+...+(1/1(2n-1)) , i.e b_n = sum (1/(2k-1)) from k=1 to n

    I need to prove that lim (b_n) as n->inf = inf
    I think I should use the squeeze theorem , but this sequence approaches infinity really slow, so I couldn't find a sequence {a_n} which approaches infinity and b_n >= a_n for almost every n...

    Any clues anyone?
    Thanks!
    1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+  \frac{1}{11}+\frac{1}{13}+....

    Notice that

    \frac{1}{5}+\frac{1}{7}>\frac{2}{6}

    \frac{1}{9}+\frac{1}{11}>\frac{2}{10}

    \frac{1}{13}+\frac{1}{15}>\frac{2}{14}

    In this way, there is a relationship between partial sums,
    leading to

    \sum_{n=1}^{\infty}\frac{1}{2n-1}>1+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+.....

    which is diverging
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  4. #4
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    Quote Originally Posted by Archie Meade View Post
    1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+  \frac{1}{11}+\frac{1}{13}+....

    Notice that

    \frac{1}{5}+\frac{1}{7}>\frac{2}{6}

    \frac{1}{9}+\frac{1}{11}>\frac{2}{10}

    \frac{1}{13}+\frac{1}{15}>\frac{2}{14}

    In this way, there is a relationship between partial sums,
    leading to

    \sum_{n=1}^{\infty}\frac{1}{2n-1}>1+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+.....

    which is diverging
    Hmm okay,
    I will think about that
    Thanks a lot!
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