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Math Help - calc 3..3 questions left NEED HELP URGENTLY

  1. #1
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    calc 3..3 questions left NEED HELP URGENTLY

    Helllo people. I have been doing this assignment and got stuck on 3 problems that I cant seem to figure out. I know its late since the assignment is due well later on today but any help will be greatly appreciated. I uploaded the section I need help with. The questions i couldnt figure out were f,k and l. For some reason my mind just hit a blank.

    Thanks in advance.
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    Hey there!

    I'm not sure if you have learned the root test for analysis yet and so I will assume that you haven't

    f) Notice that \lim_{n\rightarrow\infty}\frac{3^{2n}}{2^{3n+1}}=\  frac{9}{8}.
    This cannot possibly converge as the terms do not tend to zero as n tends to infinity. The root test would give an equal result

    k) Note that, as n becomes large, \frac{1}{\sqrt{n}+\sqrt{n-1}}\rightarrow\frac{1}{\sqrt{n}+\sqrt{n}}=\frac{1}  {2\sqrt{n}}. This is known to diverge by the integral test

    l) Again, this sequence tends to the limit 1 and so, as this is not zero, the series cannot converge

    Hope that helps!
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  3. #3
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    Quote Originally Posted by notyeteuler View Post
    Hey there!

    I'm not sure if you have learned the root test for analysis yet and so I will assume that you haven't

    f) Notice that \lim_{n\rightarrow\infty}\frac{3^{2n}}{2^{3n+1}}=\  frac{9}{8}.
    This cannot possibly converge as the terms do not tend to zero as n tends to infinity. The root test would give an equal result

    k) Note that, as n becomes large, \frac{1}{\sqrt{n}+\sqrt{n-1}}\rightarrow\frac{1}{\sqrt{n}+\sqrt{n}}=\frac{1}  {2\sqrt{n}}. This is known to diverge by the integral test

    l) Again, this sequence tends to the limit 1 and so, as this is not zero, the series cannot converge

    Hope that helps!
    The limit in f) is actually +oo, but this doesn't change the conclusion.

    For k) you'll need to set up the appropriate inequality in order to run the argument ....... \frac{1}{\sqrt{n}+\sqrt{n-1}}> \frac{1}{\sqrt{n}+\sqrt{n}}=\frac{1}{2\sqrt{n}} therefore ......
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    The limit in f) is actually +oo, but this doesn't change the conclusion.

    For k) you'll need to set up the appropriate inequality in order to run the argument ....... \frac{1}{\sqrt{n}+\sqrt{n-1}}> \frac{1}{\sqrt{n}+\sqrt{n}}=\frac{1}{2\sqrt{n}} therefore ......
    thanks for the help. I used the alternating series test and the rati test to prove the problems. I came up with the same answer. Thanks alot
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  5. #5
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    He4llo, aloufy!


    (f)\;\;\sum^{\infty}_{n=1}(-1)^{n-1}\frac{3^{2n}}{2^{3n+1}}
    Ratio test: . R \:=\:\left|\frac{a_{n+1}}{a_n}\right| \:=\:\frac{3^{2n+2}}{2^{3n+4}}\cdot\frac{2^{3n+1}}  {3^{2n}} \;=\;\frac{3^2}{2^3} \:=\:\frac{9}{8}


    \text{Since }\lim_{n\to\infty}|R| \:\geq \:1\text{, the series diverges.}




    (l)\;\;\sum^{\infty}_{n=1}\frac{4^n}{4^n + 5}
    The n^{th} term is: . a_n \:=\:\frac{4^n}{4^n + 5}

    Divide top and bottom by 4^n\!:\;\;a_n \:=\:\frac{1}{1 + \frac{5}{4^n}}

    Then: . \lim_{n\to\infty} a_n \;=\;\lim_{n\to\infty}\frac{1}{1 + \frac{5}{4^n}} \;=\;1


    \text{Since }\lim_{n\to\infty} a_n \:\neq \:0\text{, the series diverges.}

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  6. #6
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    For i:

    We could use the ratio test, but we can do one better by finding what it actually converges to.

    Rewrite as:

    2\sum_{n=1}^{\infty}\frac{1}{2^{n}}+\frac{1}{3}\su  m_{n=1}^{\infty}\left(\frac{3}{4}\right)^{n}

    We can see both of these are geometric series. We can use the formula.

    2\left(\frac{\frac{1}{2}}{1-\frac{1}{2}}\right)+\frac{1}{3}\left(\frac{\frac{3  }{4}}{1-\frac{3}{4}}\right)=3
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by aloufy View Post
    Helllo people. I have been doing this assignment and got stuck on 3 problems that I cant seem to figure out. I know its late since the assignment is due well later on today but any help will be greatly appreciated. I uploaded the section I need help with. The questions i couldnt figure out were f,k and l. For some reason my mind just hit a blank.











    Thanks in advance.
    I will, show a couple of ways of doing some of these

    a) this series can be rewritten as \frac{1}{3\cdot{2^0}}+\frac{1}{3\cdot{2^1}}+\frac{  1}{3\cdot{2^2}}+\frac{1}{3\cdot{2^3}}...+\frac{1}{  3\cdot{2^n}}

    So the series is really

    \sum_{n=0}^{\infty}\frac{1}{3\cdot{2^n}}

    This one can really be done with almost any of the tests in specific, root,ratio, limit comparison test, direct comparison test, I will opt for the LCT(limit comparison test)

    Since \lim_{n\to\infty}\frac{\frac{1}{3\cdot{2^n}}}{\fra  c{1}{2^n}}=\frac{1}{3} the series share convergence/divergence, and since \frac{1}{2^n} is a convergent geometric series both are convergent

    c)Since we can see that 2^{n+1}>2^n, \forall{n}\in\mathbb{N}
    we can see that
    \frac{1}{2^{n+1}}<\frac{1}{2^n}

    \therefore{a_{n+1}<a_n}\forall{n}\in\mathbb{N}

    and since \lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{1}{2^{  n+1}}+0 this series is conditionally convergent at least,

    Alternately seeing that \bigg|\frac{(-1)^n}{2^{n+1}}\bigg|=\frac{1}{2^{n+1}} which is a convergent geo series that this series is absolutely convergent, therefore conditionally convergent



    d) Rewriting this \sum_{n=0}^{\infty}\frac{2^{n+1}}{3^{n-3}} as \sum_{n=0}^{\infty}\frac{2\cdot{2^n}}{\frac{1}{27}  \cdot{3^n}}=54\sum_{n=0}^{\infty}\frac{2^n}{3^n}

    we can easily see this is a convergent geo series, you could have also used the ratio or root test, as well as the limit compariosn test with \frac{2^n}{3^n}


    e) Utilizing the root test we see that

    \lim_{n\to\infty}\bigg|\frac{1}{3^{\frac{n}{2}}}\b  igg|^{\frac{1}{n}}=\lim_{n\to\infty}\frac{1}{3^{\f  rac{1}{2}}}=\frac{1}{\sqrt{3}}<1

    therefore convergent, also you could use the ratio test


    f) This should be an automatic recognition of a series divergent by the n-th term test

    \lim_{n\to\infty}\frac{3^{2n}}{2^{3n+1}}=\lim_{n\t  o\infty}\frac{1}{2}\frac{9^n}{8^n}=\infty

    therefore obviously divergent


    g) This is probably most easily done by either the Direct Comparion test or LCT with \frac{1}{4n^2}

    \lim_{n\to\infty}\frac{\frac{4}{4n^2-1}}{\frac{1}{4n^2}}=4


    therefore convergent


    but you can also use the integral test, ratio test, root test

    h)THis once again should be a recogintion of nth term test

    \lim_{n\to\infty}\frac{3n^2}{(3n-1)^2}=\lim_{n\to\infty}\frac{3n^2}{9n^2-6n+1}=\frac{1}{3}\ne{0}

    therefore divergent

    EDIT: I didnt see those last four

    i) NOting that
    \sum_{n=0}^{\infty}\frac{2^{n+1}+3^{n-1}}{4^n}=2\sum_{n=0}^{\infty}\frac{2^n}{4^n}+\frac  {1}{3}\sum_{n=0}^{\infty}\frac{3^n}{4^n}
    so since this series is the sum of two convergent geo series it itself is convergent


    j) Seeing that
    \sum_{n=0}^{\infty}\ln(4n+1)-\ln(2n+1)=\sum_{n=0}^{\infty}\ln\bigg(\frac{4n+1}{  2n+1}\bigg)
    now n-th term test should be screaming

    let
    L=\lim_{n\to\infty}\ln\bigg(\frac{4n+1}{2n+1}\bigg  )\Rightarrow{e^L=\lim_{n\to\infty}\frac{4n+1}{2n+1  }=2}
    so
    e^L=2\Rightarrow{L=\ln(2)\ne{0}}

    therefore this series is divergent


    k) YOu can either use LCT with \frac{1}{2\sqrt{n}} or see that

    \exists{N}\backepsilon\forall{n>N}\frac{1}{\sqrt{n  }+\sqrt{n+1}}>\frac{1}{2\sqrt{n}}

    and sicne the right side is a divergent p-series this series diverges by the Direct comparison tset


    l) This should scream n-th term test as well

    \lim_{n\to\infty}\frac{4^n}{4^n+5}=1\ne{0}

    and just in case you are rusty on limits
    \lim_{n\to\infty}\frac{4^n}{4^n+5}=\lim_{n\to\inft  y}\frac{4^n}{4^n+5}\cdot\frac{\frac{1}{4^n}}{\frac  {1}{4^n}}=\lim_{n\to\infty}\frac{1}{1+\frac{5}{4^n  }}=\frac{1}{1+0}\ne{0}


    therefore divergent
    Last edited by Mathstud28; May 30th 2008 at 12:07 PM.
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