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Testing for Convergence

  1. Apr 6, 2004 #1
    We just learned Alternating Series, but I am still unsure of the procedure. A couple of problems in the book have stumped me, and I need to test them for convergence.

    1. Sum n=1 going to infinity of [cos(n*pi) / (n^(7/8))]
    2. Sum n=1 going to infinity of (-n/5)^n

    I am really lost, I'd really appreciate the help...
     
  2. jcsd
  3. Apr 6, 2004 #2
    Okay, an alternating series converges if the absolute value of the terms in it approaches zero. It's basically as if you take a big step forwards, a smaller step back, an even smaller step forwards and so on. Try it yourself (but not in class or you'll get funny looks).

    1) cos(n * pi) is 1 for n even and -1 for n odd. So your series is basically just

    -1/(1^7/8) + 1/(2^7/8) -1/(3^7/8) + ...

    All you have to do now is show that as n->infinity, 1/(n^7/8) -> 0, and you should have no problem with that.

    2) Just rewrite this as ((-1)^n)/(5^n) and observe that (-1)^n is positive for even n and negative for odd n, so you have an alternating series. Then this question is pretty much the same as the last one.
     
  4. Apr 6, 2004 #3
    so for #1, i should check the lim as n -> infinity of bn and see if it equals 0? and bn in this case is [cos(n*pi) / (n^(7/8))] ... cause according to the alternating series theorem, if the limit = 0, then it is convergent...

    havent gotten to #2 yet. stay tuned ;)
     
  5. Apr 6, 2004 #4
    Yeah, that's pretty much right. For these questions you need to a) show they're alternating and b) the limit as n -> inf is zero.
     
  6. Apr 6, 2004 #5
    ok, did them, lets see if they're right.

    1) lim n->inf [cos(n*pi) / (n^(7/8))] = 1/inf == 0 therefore, the Sum is convergent

    2) lim n->inf 5^(-n) == 0 and therefore, the Sum is also convergent

    yes? no? thanks for your help
     
  7. Apr 6, 2004 #6
    Bingo.

    Wasn't as hard as all that cos(n*pi) stuff made it look, eh? Just be glad they didn't ask you what the thing actully converged to; that's much harder.
     
  8. Apr 6, 2004 #7
    oh i know, thanks again for the help. really helped alot. i'm always pleased with the helpfulness this site brings.
     
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