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Root test & Ratio test

  1. Mar 8, 2010 #1
    Let ∑ak be a series with positive terms.
    Ratio test:
    Suppose ak+1/ak -> c.
    If c<1, then ∑ak converges.
    If c>1, then ∑ak diverges.
    If c=1, the test is inconclusive.

    What if ak+1/ak diverges (i.e. ak+1/ak->∞)? Do we count this as falling into the case c>1? Can we say whether ∑ak converges or not?

    Root test:
    Suppose limsup (ak)1/k = r.
    If r<1, then ∑ak converges.
    If r>1, then ∑ak diverges.
    If r=1, the test is inconclusive.

    What if limsup (ak)1/k = ∞? Do we count this as falling into the case r>1? Can we say whether ∑ak converges or not?


    Thanks for clarifying!
     
  2. jcsd
  3. Mar 8, 2010 #2
    Just unwind the definitions, and it should be clear.

    First question: if [tex]a_{k+1}/a_k\to\infty[/tex], then for M as big as you like, there's some N such that [tex]a_{k+1}/a_k>M[/tex] for all k>N. That means that for k>N, we have [tex]|a_k|>M^{k-N}[/tex]. Clearly that's diverging.

    Second question: if [tex]\text{lim sup }a_k^{1/n}=\infty[/tex], then for M and N as large as you'd like, there's a k>N with [tex]a_k^{1/k}>M[/tex], i.e. [tex]|a_k|>M^k[/tex]. Again, clearly divergent.
     
  4. Mar 9, 2010 #3
    So in the ratio test, the case c=∞ is allowed?

    c=∞>1 => ∑ak diverges?
     
  5. Mar 9, 2010 #4
    "The" ratio test requires that the sequence of ratios converges--it's right there in the definition you gave. If the sequence diverges to infinity, though, a simple modification of the argument yields the same result.

    Whether that distinction is meaningful is up to you.
     
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