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Sequences and series

  1. Sep 20, 2010 #1
    Struggling with this topic!! :(
    got a couple of questions.

    1. The problem statement, all variables and given/known data
    1) Determine the value of the improper integral when using the integral test to show that
    [tex]\sum[/tex]k/(e^k/5) is convergant
    given answers are
    a)50/e
    b)-1/(5e^1/5)
    c)5
    d)5e
    e)1/50e

    2) determine whether [tex]\sum[/tex] (n+5)/(n^3-2n+3) is convergant or divergant

    3) find interval of convergance of the series [tex]\sum[/tex]5(x-3)^n


    3. The attempt at a solution

    1) by integral test, f(x) = k/(e^k/5) = ke^(-k/5)
    and the integral of that is [-1/5*ke^(-k/5)] -1/5*e^(-k/5).. thats integration by parts. so thats the imporper integral???
    i dunno how to relate that to any of the answers

    2) i multiplied top and bottom by n^2 and then limit of 1/n and all varieties is 0 so i end up with lim 0+0/n-0+0=0 so converges... i just needa check if thats right?

    3) i have nothing in my textbooks about finding intevals. i dunno where to even start.


    thanks for any help
     
  2. jcsd
  3. Sep 20, 2010 #2
    1) What is the initial index of the sum? That should determine the lower limit of integration for the corresponding improper integral. Also there are multiple errors in your integration by parts (including not bothering to write out the integral sign for the second term). My advice is to make the substitution u = x/5 to avoid confusion.

    2) No, showing that the limit of the terms is 0 does not show that the series converges. Use the limit comparison test.

    3) Use the ratio test.
     
  4. Sep 20, 2010 #3
    ill try those tests for 2 and 3... part one i got from a similar answer int he textbook... fromt he answers textbook.... it shows the integral of xe^-x.dx to be (-xe^-x) - (e^-x) so i just wnet off that subbing in the 1/5 fraction. i really have no idea how to integrate that. it just says in brackets that its done by parts. that function given in the question is continuous on interval [1,infinity) which makes the lower limit 1 yes? this questions got me so lost. dont stress much about the correct terminology for the integration. i can work all that out. i just kinda need to know which part of the solution is the imporper integral. i thought it was just the integral of f(x) when f(x) = an??
     
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