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!

Calculus 2 - Infinite Series

  1. Nov 4, 2011 #1
    This is a similar question that I have to the other one that I posted.

    Determine if the following series converges or diverges.

    sum[k=1,inf] 3/(k(k+3))

    I applied the limit comparison test with 1/k^2
    also k(k+3))=k^2+3k

    lim k->inf [ 3/(k^2+3k) ]/[ 1/k^2 ] = lim k->inf (3k^2/(k^2+3k) =H 3

    because sum[k=1,inf] 1/k^2 is a convergent p-series than sum[k=1,inf] 3/(k(k+3)) must also converge by the limit comparison test.

    I plugged this into wolfram alpha
    sum[n=1,inf] of 3/(k(k+3))
    and it said that the sum does not converge by the limit comparison test. Am I doing something wrong?

    Thanks for any help.
     
  2. jcsd
  3. Nov 4, 2011 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Never trust Wolfram Alpha. Reagan said "trust but verify". Plug this into WA. "sum 3/(k*(k+3)) for k from 1 to infinity". You managed to confuse the poor thing somehow. Probably the [n=1,inf] part again.
     
    Last edited: Nov 4, 2011
  4. Nov 4, 2011 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Wolfram Alpha keeps giving you wrong answers (on this and some other problems). What would you conclude from that?

    RGV
     
  5. Nov 4, 2011 #4
    Oh I screwed up the indexes. Entering that actually gives me the right answer. Thanks. You can conclude that woflram alpha isn't the be all end all super calculator.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Calculus 2 - Infinite Series
Loading...