## Ratio Test for Series Proof

1. The problem statement, all variables and given/known data
Show that if $$Lim|\frac{a_{n+1}}{a_{n}}| = L > 1,$$ then $${a_{n}\rightarrow \infty$$ as $$n\rightarrow\infty$$

Also, from that, deduce that $$a_{n}$$ does not approach 0 as $$n \rightarrow \infty$$.

2. Relevant equations
The book suggests showing some number r>1 such that for some number N, $$|a_{n+1}|> r|a_{n}|$$ for all n >N.
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 Recognitions: Gold Member Science Advisor Staff Emeritus Okay, and what have you done on this problem? Have you shown, perhaps by induction on n, that $|a_{n+1}|> r|a_n|$? Once you've done that, you might consider the "comparison test".
 How would I start that proof by induction? How can I verify that $|a_{2}|> r|a_{1}|$. Also, for the second part, once I show that $$|a_{n}|$$ tends to $$\infty$$ isn't it basic logic that $$a_{n}$$ cannot approach 0?

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## Ratio Test for Series Proof

 Quote by dtl42 How would I start that proof by induction? How can I verify that $|a_{2}|> r|a_{1}|$.
You can't. It's not necessarily true. However, since the limit $a_{n+1}/a_n$ 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 $$|a_{n}|$$ tends to $$\infty$$ isn't it basic logic that $$a_{n}$$ cannot approach 0?
Yes, it is. Just state the basic logic.

 Quote by HallsofIvy You can't. It's not necessarily true. However, since the limit $a_{n+1}/a_n$ 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, $|a_{2}|>r|a_{1}|$?

 Quote by dtl42 Do you mean greater than 1, or am I really missing something? And how would I start the induction? Just that for some N, $|a_{2}|>r|a_{1}|$?
If L>1, then is the sequence $${a_{n}}$$ bounded or unbounded?
Suppose not, If L<1, then what happens when $${\lim }\limits_{n \to \infty } a_{n}$$?
 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?

 Quote by dtl42 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 $$|a_{n+1}|> r|a_{n}|$$ for all indices $$n\geq N$$, you can use the Binomial formula to show that the sequences is unbounded. Hope that's clear.
 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?

 Quote by oceanflavored 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?
 Quote by konthelion Correct. Now, since book suggested: show that $$|a_{n+1}|> r|a_{n}|$$ for all indices $$n\geq N$$, you can use the Binomial formula to show that the sequences is unbounded. Hope that's clear.
 Well, you can use Bernoulli's Inequality, which is $$(1+b)^n \geq 1 +nb$$ Suppose that $$L>1$$, then define $$b=\frac{L+1}{2}$$ since b