Calculus II

Marie1309

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Mar 22, 2019
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When we use the ratio test, we find the limit:

\(\displaystyle L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\)

The ratio test states that:
  • if \(L<1\) then the series converges absolutely;
  • if \(L<1\) then the series is divergent;
  • if \(L=1\) or the limit fails to exist, then the test is inconclusive, because there exist both convergent and divergent series that satisfy this case.
And so, because we require \(L<1\), the interval of convergence is necessarily open, that is, we do not include the end-points.
Thank you so much. But you put L<1 two times
 

Jomo

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Dec 30, 2014
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3,326
MarkFL meant to write it is divergent if L>1

Recall that a necessary condition for convergence is that an-->0 as n--oo
Now if L>1, then an+1> an and therefore the (positive) terms are getting bigger and hence not approaching 0. So it diverges
That's how I remember the rules!
 
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