Root test & Ratio test

  • Thread starter kingwinner
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  • #1
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Main Question or Discussion Point

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!
 

Answers and Replies

  • #2
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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.
 
  • #3
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So in the ratio test, the case c=∞ is allowed?

c=∞>1 => ∑ak diverges?
 
  • #4
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So in the ratio test, the case c=∞ is allowed?
"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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