Proof of the Ratio Test and the Triangle Inequality

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SUMMARY

The discussion centers on proving that if the limit inferior of the absolute ratio of consecutive terms, lim inf |ak+1/ak| > 1, then the series Σ ak diverges. The user outlines a proof strategy involving the existence of a constant r such that lim inf |ak+1/ak| > r > 1, leading to the conclusion that Σ |an| * r^k diverges as a geometric series. The user seeks clarification on transitioning from the divergence of absolute values to the divergence of the series itself, particularly regarding conditional convergence.

PREREQUISITES
  • Understanding of limit inferior in sequences
  • Familiarity with the Ratio Test for series convergence
  • Knowledge of geometric series and their convergence properties
  • Concept of absolute and conditional convergence in series
NEXT STEPS
  • Study the Ratio Test and its applications in series convergence
  • Explore the concept of absolute versus conditional convergence in detail
  • Learn about the comparison test for series divergence
  • Investigate geometric series and their convergence criteria
USEFUL FOR

Mathematics students, particularly those studying real analysis or series convergence, as well as educators looking for examples of the Ratio Test and its implications.

Fiz2007
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Homework Statement



Prove: If the limit inf as k goes to infinity of abs(ak+1 / ak) > 1 then the sum from 1 to infinity of ak diverges


Homework Equations




The Attempt at a Solution



So far I have this:

Suppose lim inf abs(ak+1/ak) >1
then, there exists an r such that lim inf abs(ak+1/ak) > r > 1
then [tex]\exists[/tex] N an element of the natural numbers such that k >= N implies
abs(ak+1/ak) > r
that is, for k >= N, abs(ak+1) > abs(ak) r
and, abs (an+1) > r abs(an)
abs (an+2) > r abs (an+1) > r^2 abs(an)
and in general, abs (an+k) > r^k abs(an)

the series the sum from 1 to infinity of abs(an) * r^k diverges to infinity (geometric series with r>1).

Therefore the sum from 1 to infinity of abs(an+k) diverges to infinity by the comparison test.

Then I'm stuck. I don't know how to get from the absolute value diverging to the series diverging. It obvioulsy does not converge absolutely, but what about conditional convergence. Please help!

Thanks
 
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Fiz2007 said:

Homework Statement



Prove: If the limit inf as k goes to infinity of abs(ak+1 / ak) > 1 then the sum from 1 to infinity of ak diverges


Homework Equations




The Attempt at a Solution



So far I have this:

Suppose lim inf abs(ak+1/ak) >1
then, there exists an r such that lim inf abs(ak+1/ak) > r > 1
then [tex]\exists[/tex] N an element of the natural numbers such that k >= N implies
abs(ak+1/ak) > r
that is, for k >= N, abs(ak+1) > abs(ak) r
and, abs (an+1) > r abs(an)
abs (an+2) > r abs (an+1) > r^2 abs(an)
and in general, abs (an+k) > r^k abs(an)

the series the sum from 1 to infinity of abs(an) * r^k diverges to infinity (geometric series with r>1).

Therefore the sum from 1 to infinity of abs(an+k) diverges to infinity by the comparison test.

Then I'm stuck. I don't know how to get from the absolute value diverging to the series diverging. It obvioulsy does not converge absolutely, but what about conditional convergence. Please help!

Thanks

You've shown the absolute values of the terms in your series are bounded below by a geometric series whose terms go to infinity. The terms in a series have to go to zero if it will have any hope of converging.
 

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