1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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]


    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

    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.

    So it looks as tho it converges to 0 as n tends to infinity.

    tan^{-1}(n) = \frac{\pi}{4},1.107,1.25...
    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.

    Again looks like it converges to 0 as n tends to infinity.

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

    \sum_{n=1}^{99} (-1)^n=1-1+1-1+1-1
    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.

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

    \sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}}=1+0.353+0.192+.....
    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.

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


    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


    Staff: Mentor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook