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Baby Rudin, Example 3.35

  • Thread starter union68
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Homework Statement



Consider the series

[tex]\frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \ldots[/tex].

He then proceeds to write

[tex]\lim \textnormal{inf}_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \left(\frac{2}{3}\right)^n = 0[/tex]

and

[tex]\lim\textnormal{sup}_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{1}{2} \left(\frac{3}{2}\right)^n = +\infty[/tex].

This is in reference to the ratio test, by the way.

Homework Equations





The Attempt at a Solution



How did he arrive at these calculations? If for the lower limit we select [itex]a_{n+1} = 1/3^n[/itex] and [itex]a_n = 1/2^n[/itex], then we can recover his answer. But

[tex]\left\{\frac{a_{n+1}}{a_n}\right\} = \left\{\left(\frac{2}{3}\right)^n\right\}[/tex]

is a convergent sequence and the upper and lower limits must be equal, correct? How is the upper limit [itex]\infty[/itex]?

On the other hand, if we select [itex]a_{n+1} = 1/2^{n+1}[/itex] and [itex]a_n = 1/3^n[/itex], then we can recover his answer for the upper limit.

Isn't there some ambiguity with selecting [itex]a_{n+1}[/itex] and [itex]a_n[/itex] from this series?
 

Answers and Replies

  • #2
jbunniii
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[tex]\frac{a_{n+1}}{a_n} = \left(\frac{2}{3}\right)^n[/tex]

is true for odd [itex]n[/itex] only.

[tex]\frac{a_{n+1}}{a_n} = \frac{1}{2} \left(\frac{3}{2}\right)^n[/tex]

is true for even [itex]n[/itex] only.

(or it might be the other way around, depending on how you index the terms)

Thus

[tex]\frac{a_{n+1}}{a_n}[/tex]

ping-pongs back and forth between the larger subsequence and the smaller subsequence. The larger one produces the lim sup and the smaller one produces the lim inf. Since the lim sup and the lim inf are unequal, it follows that

[tex]\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}[/tex]

does not exist.
 
Last edited:
  • #3
140
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Ah crap, got it! I should have wrote the sequence out so I could keep my indexing strait.

Thanks very much.
 

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