Results 1 to 11 of 11

Math Help - is this correct,please help confirm

  1. #1
    Member
    Joined
    Feb 2009
    Posts
    130

    is this correct,please help confirm

    Determine the series is absolutely convergent or diverges by using ratio test.
    b) \sum ^\infty _{n=1} {(1+\frac{1}{n})}^{n^2}
    \frac {{(1+\frac{1}{n+1})}^{(n+1)^2}}{{(1+\frac{1}{n})}^  {n^2}}
    = \frac {{(1+\frac{1}{n+1})}^{n^2}+{(1+\frac{1}{n+1})}^{2n  }+{(1+\frac{1}{n+1})}^{2}}{{(1+\frac{1}{n})}^{n^2}  }

    = \frac{e^2+e^ne^n+1}{e^2}

    is that the final answer? can it be simplified any further?

    how do i simplifly this question,please help. really appreaciate all your help & support
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by anderson View Post
    Determine the series is absolutely convergent or diverges by using ratio test.
    b) \sum ^\infty _{n=1} {(1+\frac{1}{n})}^{n^2}
    \frac {{(1+\frac{1}{n+1})}^{(n+1)^2}}{{(1+\frac{1}{n})}^  {n^2}}
    = \frac {{(1+\frac{1}{n+1})}^{n^2}+{(1+\frac{1}{n+1})}^{2n  }+{(1+\frac{1}{n+1})}^{2}}{{(1+\frac{1}{n})}^{n^2}  }

    = \frac{e^2+e^ne^n+1}{e^2}

    is that the final answer? can it be simplified any further?

    how do i simplifly this question,please help. really appreaciate all your help & support
    There are a number of errors, including:

    1. The numerator of the second line of working should be a product not a sum.

    2. \lim_{n \to + \infty} \left(1+\frac{1}{n+1}\right)^{2n} \neq e^n e^n.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2009
    Posts
    130
    Determine the series is absolutely convergent or diverges by using ratio test.
    b) \sum ^\infty _{n=1} {(1+\frac{1}{n})}^{n^2}
    \frac {{(1+\frac{1}{n+1})}^{(n^2+2n+1)}}{{(1+\frac{1}{n}  )}^{2}}
    \frac {{e}^{(n^2+2n+1)}}{{e}^{n^2}}

    i dont know how should i continue from here ..need some help & guidance for this question...for all i know the final answer is e, am wondering how to get the answer.

    appreciate all your help & support.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    Quote Originally Posted by anderson View Post
    Determine the series is absolutely convergent or diverges by using ratio test.
    b) \sum ^\infty _{n=1} {(1+\frac{1}{n})}^{n^2}
    \frac {{(1+\frac{1}{n+1})}^{(n+1)^2}}{{(1+\frac{1}{n})}^  {n^2}}
    = \frac {{(1+\frac{1}{n+1})}^{n^2}+{(1+\frac{1}{n+1})}^{2n  }+{(1+\frac{1}{n+1})}^{2}}{{(1+\frac{1}{n})}^{n^2}  }

    = \frac{e^2+e^ne^n+1}{e^2}

    is that the final answer? can it be simplified any further?

    how do i simplifly this question,please help. really appreaciate all your help & support
    (1+\frac{1}{n})^{n^2}=\left((1+\frac{1}{n})^{n}\ri  ght)^n.
    Apply the Root Test.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Feb 2009
    Posts
    130
    thank you for replying, not sure about the root test.can anyoine help me.
    thank you for all your help & support.

    i think i got the answer:
    <br /> <br />
\frac {{(1+\frac{1}{n+1})}^{n^2}{(1+\frac{1}{n+1})}^{2n}  {(1+\frac{1}{n+1})}^{2}}{{(1+\frac{1}{n})}^{n^2}}<br />
    {(1+\frac{1}{n+1})}^{2n}{(1+\frac{1}{n+1})}^{2}
    =(e)(1)
    =e

    is it correct? please help confirm,thank you for all your help & support.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    Quote Originally Posted by anderson View Post
    thank you for replying, not sure about the root test.can anyoine help me.
    thank you for all your help & support.

    i think i got the answer:
    <br /> <br />
\frac {{(1+\frac{1}{n+1})}^{n^2}{\color{green}{(1+\frac{  1}{n+1})}^{2n}}{\color{red}{(1+\frac{1}{n+1})}^{2}  }}{{(1+\frac{1}{n})}^{n^2}}<br />
    {(1+\frac{1}{n+1})}^{2n}{(1+\frac{1}{n+1})}^{2}
    =(e)(1)
    =e

    is it correct? please help confirm,thank you for all your help & support.
    Well.
    Sorry, I did not follow your solution. show work please.
    the green one \rightarrow e^2
    the red one \rightarrow 1 as n\rightarrow\infty.
    so we can ignore the red one.
    Now, we are left with \lim_{n\to\infty} \frac{( 1 + \frac{1}{n+1})^{n^2} }{ (1 + \frac{1}{n})^{n^2}}
    Try to evaluate it, then multiply the result by e^2 to get the answer of the limit of the Ratio Test.

    Actually, I did not know what are trying to use the Ratio Test.
    It can be tested for convergence/divergence easily.
    Clearly \left( 1 + \frac{1}{n} \right)^{n^2} \rightarrow \infty as n \rightarrow \infty.
    so the series diverges by the Test for Divergence (the nth term test).
    Or note that:
    \left( 1 + \frac{1}{n} \right)^{n^2} \geq \left( 1 + \frac{1}{n} \right)^{n}.
    Then, the Basic Comparison Test will finish it.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Feb 2009
    Posts
    130
    hi General

    thank you for replying.

    this is the question:
    Determine the series is absolutely convergent or diverges by using ratio test.
    b) \sum ^\infty _{n=1} {(1+\frac{1}{n})}^{n^2}
    \frac {{(1+\frac{1}{n+1})}^{(n+1)^2}}{{(1+\frac{1}{n})}^  {n^2}}
    = \lim_{n\to\infty} \frac {{(1+\frac{1}{n+1})}^{n^2}{(1+\frac{1}{n+1})}^{2n}  {(1+\frac{1}{n+1})}^{2}}{{(1+\frac{1}{n})}^{n^2}}
    = \lim_{n\to\infty} {(1+\frac{1}{n+1})}^{2n}{(1+\frac{1}{n+1})}^{2}
    = \lim_{n\to\infty} (e)(1)
    =e

    is it correct? can someone help to confirm?

    thank you in advance for all your help & guidance.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    Quote Originally Posted by anderson View Post
    hi General

    thank you for replying.

    this is the question:
    Determine the series is absolutely convergent or diverges by using ratio test.
    b) \sum ^\infty _{n=1} {(1+\frac{1}{n})}^{n^2}
    \frac {{(1+\frac{1}{n+1})}^{(n+1)^2}}{{(1+\frac{1}{n})}^  {n^2}}
    = \lim_{n\to\infty} \frac {{(1+\frac{1}{n+1})}^{n^2}{(1+\frac{1}{n+1})}^{2n}  {(1+\frac{1}{n+1})}^{2}}{{(1+\frac{1}{n})}^{n^2}}
    = \lim_{n\to\infty} {(1+\frac{1}{n+1})}^{2n}{(1+\frac{1}{n+1})}^{2}
    = \lim_{n\to\infty} (e)(1)
    =e

    is it correct? can someone help to confirm?

    thank you in advance for all your help & guidance.
    Wrong.
    You canceled two things. which is wrong. they are not equal.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Feb 2009
    Posts
    130
    hi General

    thank you for replying.you are right. is this the correct working.

    this is the question:
    Determine the series is absolutely convergent or diverges by using ratio test.
    b) \sum ^\infty _{n=1} {(1+\frac{1}{n})}^{n^2}
    \frac {{(1+\frac{1}{n+1})}^{(n+1)^2}}{{(1+\frac{1}{n})}^  {n^2}}
    = \lim_{n\to\infty} \frac {{(1+\frac{1}{n+1})}^{n^2}{(1+\frac{1}{n+1})}^{2n}  {(1+\frac{1}{n+1})}^{2}}{{(1+\frac{1}{n})}^{n^2}}
    = \lim_{n\to\infty} \frac{(e^2)(e)(1)}{e^2}
    =e
    e>1, divergent.
    thank you for all your help & guidance.really appreciate.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    The limit = e, Right.
    But sorry, am not sure if you really can cancel them or not.

    See:
    http://www.wolframalpha.com/input/?i...es+to+infinity
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Feb 2009
    Posts
    130
    hi General

    thank you for helping to confirm. i'm not sure about that either, but i dont know any other way to get the answer,e.

    can anyone help to confirm this doubt? really appreciate all your help & contribution.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 15
    Last Post: July 29th 2011, 02:39 AM
  2. Please confirm this for me :)
    Posted in the Algebra Forum
    Replies: 6
    Last Post: February 20th 2010, 05:46 AM
  3. can someone confirm this
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 30th 2009, 10:36 AM
  4. can someone confirm that this is right
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 27th 2009, 12:11 PM
  5. Composite & Quotient Rule -confirm correct answer?
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: June 9th 2009, 04:23 PM

Search Tags


/mathhelpforum @mathhelpforum