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Ratio Test for Series Proof

  • Thread starter dtl42
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Homework Statement


Show that if [tex] Lim|\frac{a_{n+1}}{a_{n}}| = L > 1, [/tex] then [tex]{a_{n}\rightarrow \infty[/tex] as [tex]n\rightarrow\infty [/tex]

Also, from that, deduce that [tex]a_{n}[/tex] does not approach 0 as [tex]n \rightarrow \infty [/tex].

Homework Equations


The book suggests showing some number r>1 such that for some number N, [tex]|a_{n+1}|> r|a_{n}|[/tex] for all n >N.
 

Answers and Replies

  • #2
HallsofIvy
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Okay, and what have you done on this problem? Have you shown, perhaps by induction on n, that [itex]|a_{n+1}|> r|a_n|[/itex]? Once you've done that, you might consider the "comparison test".
 
  • #3
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How would I start that proof by induction? How can I verify that [itex]
|a_{2}|> r|a_{1}|
[/itex]. Also, for the second part, once I show that [tex]|a_{n}|[/tex] tends to [tex]\infty[/tex] isn't it basic logic that [tex]a_{n}[/tex] cannot approach 0?
 
  • #4
HallsofIvy
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How would I start that proof by induction? How can I verify that [itex]
|a_{2}|> r|a_{1}|
[/itex].
You can't. It's not necessarily true. However, since the limit [itex]a_{n+1}/a_n[/itex] is less than 1, it must be true for some N. Start your induction from that.

Also, for the second part, once I show that [tex]|a_{n}|[/tex] tends to [tex]\infty[/tex]
isn't it basic logic that [tex]a_{n}[/tex] cannot approach 0?
Yes, it is. Just state the basic logic.
 
  • #5
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You can't. It's not necessarily true. However, since the limit [itex]a_{n+1}/a_n[/itex] is less than 1, it must be true for some N. Start your induction from that.
Do you mean greater than 1, or am I really missing something? And how would I start the induction? Just that for some N, [itex]|a_{2}|>r|a_{1}|[/itex]?
 
  • #6
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Do you mean greater than 1, or am I really missing something? And how would I start the induction? Just that for some N, [itex]|a_{2}|>r|a_{1}|[/itex]?
If L>1, then is the sequence [tex]{a_{n}}[/tex] bounded or unbounded?
Suppose not, If L<1, then what happens when [tex]{\lim }\limits_{n \to \infty } a_{n}[/tex]?
 
  • #7
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If L>1 then the sequence would be unbounded right? Because the next larger term is always of a greater magnitude than the previous. If L is less than 1, then the sequence is bounded, and the limit goes to 0?
 
  • #8
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If L>1 then the sequence would be unbounded right? Because the next larger term is always of a greater magnitude than the previous. If L is less than 1, then the sequence is bounded, and the limit goes to 0?
Correct.

Now, since book suggested: show that [tex] |a_{n+1}|> r|a_{n}|[/tex] for all indices [tex]n\geq N[/tex], you can use the Binomial formula to show that the sequences is unbounded. Hope that's clear.
 
  • #9
REALLY IRRELEVANT but....
i've always wondered this, but how do you guys get all those math symbols in there? like the absolute value symbol, or the greater than equal to sign?
 
  • #10
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  • #11
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Correct.

Now, since book suggested: show that [tex] |a_{n+1}|> r|a_{n}|[/tex] for all indices [tex]n\geq N[/tex], you can use the Binomial formula to show that the sequences is unbounded. Hope that's clear.
What do you mean by the Binomial Formula? I'm still kind of confused after taking several days away.
 
  • #12
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Well, you can use Bernoulli's Inequality, which is [tex](1+b)^n \geq 1 +nb[/tex]

Suppose that [tex] L>1[/tex], then define [tex]b=\frac{L+1}{2}[/tex] since b<L. There exists a natural number N such that
[tex]\frac{a_{n+1}}{a_{n}} \geq b[/tex] for all indices [tex]n \geq N[/tex] (just a reiteration of the problem)

From here, use the Bernoulli's inequality to show that for some k and let r = [tex]b^k[/tex], then [tex]|a_{N+k}| \geq r|a_{N}|[/tex] which implies that the sequence [tex]a_{n}[/tex](the hint that your book gave) is unbounded.
 

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