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Sequences, Series, Convergence and Divergence

  1. Oct 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Q1 Are the following sequences divergent or convergent as n tends to infinity.

    a: [itex]\frac{5n+2}{n-1}[/itex]

    b: [itex]tan^{-1}(n)[/itex]

    c:[itex]\frac{2^n}{n!}[/itex]

    Q2 Evaluate:....

    a: [itex]\sum_{n=1}^{\infty} 3^{\frac{n}{2}} [/itex]

    b: [itex]\sum_{n=1}^{99} (-1)^n [/itex]

    Q3 Find whether the following converge or diverge

    a:[itex]\sum_{n=1}^{\infty} \frac{n-1}{n}[/itex]

    b:[itex]\sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}}[/itex]




    2. Relevant equations
    |\frac{a_{n+1}}{a_n}|


    3. The attempt at a solution
    Most of these I have no clue on how to format it mathmatically correct but I have given it my best shot, Id be surprised if I have any correct mind you.

    Q1a
    [tex]
    \frac{5n+2}{n-1}=12,\frac{17}{2},\frac{22}{3},\frac{27}{4}...\\
    [/tex]
    So it looks as tho it converges to 0 as n tends to infinity.

    Q1b
    [tex]
    tan^{-1}(n) = \frac{\pi}{4},1.107,1.25...
    [/tex]
    From messing on the calculator I can see that it tends to pi/2 as n tends to infinity but dont know how to show it mathmatically.

    Q1c
    [tex]
    \frac{2^n}{n!}=2,2,1.33,0.66,0.266
    [/tex]
    Again looks like it converges to 0 as n tends to infinity.

    Q2a
    [tex]
    \sum_{n=1}^{\infty} 3^{\frac{n}{2}}=1+\frac{1}{\sqrt{3}}+\frac{1}{3}+0.192+\frac{1}{9}
    [/tex]
    This looks as though the limit is 2, as n tends to infinity

    Q2b
    [tex]
    \sum_{n=1}^{99} (-1)^n=1-1+1-1+1-1
    [/tex]
    I noted that when n is even it is +1 and when it is odd it is -1 and since 99 is odd, then the limit is 0 as n tends to 99.

    Q3a
    [tex]\sum_{n=1}^{\infty} \frac{n-1}{n}=\frac{\frac{n}{n}-\frac{1}{n}}{\frac{n}{n}}=1-\frac{1}{n} \\
    =1-1+1-\frac{1}{2}+1-\frac{1}{3}+1-\frac{1}{4}...[/tex]
    here it seems it diverges to infinity as n tends to infinity (due to the +1's)

    Q3b
    [tex]
    \sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}}=1+0.353+0.192+.....
    [/tex]
    Here is looks like it converges to the limit of 2 as n tends to infinity.


    Sorry for the mass of questions, I am not sure about any of them so any advice would be much appreciated.

    Thanks.
     
    Last edited: Oct 13, 2013
  2. jcsd
  3. Oct 13, 2013 #2

    joshmccraney

    User Avatar
    Gold Member

    hey! sorry, i didn't look through all the equations, but i'll offer some help on [itex]1c)[/itex] [tex]\frac{2^n}{n!}=\frac{2}{n}\cdot\frac{2}{(n-1)}\cdot\frac{2}{(n-2)}\cdot \cdot \cdot \cdot \frac{2}{3}\cdot\frac{2}{2}\cdot\frac{2}{1}[/tex]. now what can we do (compare)...i'll let you think on this.
     
  4. Oct 13, 2013 #3

    Mark44

    Staff: Mentor

    FaraDazed,
    Please limit the number of problems you post in a thread to one or, at most, two.

    I have closed this thread. Feel free to start new threads with a problem or two in each.
     
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