Question about one part of the Ratio Test proof

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The discussion centers on the proof of the Ratio Test in calculus, specifically the transition from the inequality |a_{N+1}| < r |a_N| to |a_{N+2}| < r^2 |a_N|. The user Ivan77 seeks clarification on how the exponent of r increases with each term in the sequence. The explanation provided clarifies that by substituting |a_{N+1}| into the inequality for |a_{N+2}|, one can see that the pattern continues, leading to |a_{N+k}| < r^k |a_N|. This understanding is crucial for grasping the generalization of the Ratio Test.

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ivan77
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Hi,

there is a proof of the ratio test that I have seen a couple of times here:

http://tutorial.math.lamar.edu/Classes/CalcII/RatioTest.aspxI don't know how to use latex, so abs(x) will mean absolute value

I am ok with this:

abs( aN+1) < r abs(aN)
I don't understand where the r2 come from in the next line
abs( aN+2) < r abs(aN+1)< r2 abs(aN)

Similarly, I don't understand how to get to the generalization of
abs( aN+k) < rk abs(aN+1)I should probably add that I understand all other aspects (i.e. using comparison test), but since I don't understand this part, my understanding of the proof fails.

Thanks,

Ivan77
 
Last edited:
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Just in case your were wondering, you don't need LaTeX for absolute value, the key is right above your Enter key on your keyboard (PC): |

So we know \left|\frac{a_{n+1}}{a_n}\right| &lt; r which is the same as |a_{n+1}| &lt; r |a_n|. This is saying that the n+1st term (the term after the nth term) in the sequence is the nth term multiplied by r.

What they are doing in the proof is going backwards. Think of it this way, it might help you better visualize it:

Starting with |a_{N+1}| &lt; r |a_N| and |a_{N+2}| &lt; r |a_{N+1}| we can substitute for |a_{N+1}| below:

<br /> \begin{eqnarray*}<br /> |a_{N+2}| &amp;&lt;&amp; r |a_{N+1}|\\<br /> &amp;&lt;&amp; r (r |a_{N}|) = r^2 |a_N|<br /> \end{eqnarray*}<br />

And you can continue this process

<br /> \begin{eqnarray*}<br /> |a_{N+3}| &amp;&lt;&amp; r |a_{N+2}|\\<br /> &amp;&lt;&amp; r (r^2 |a_{N}|) = r^3 |a_N|<br /> \end{eqnarray*}<br />


Does that help or is it still confusing?
 
Hugely appreciate the reply scurty! It answered my question. I've been staring at this proof for a couple of hours.

I was getting stuck on the logic of how we would know that the r^2 would not make the term less than the An+1 rather that using the inequality given.

This is what I was gettign stuck on:

r|aN+1| < r(r|aN|)

has to be true since

since
|aN+1| < r|aN|

and multiplying by a factor of r will not change that fact.

Why couldn't I see that simply distributing the r in the right term would answer the question?
The odd thing about my self studying calculus is that the concepts as well answering questions/application is not a problem, but now and then, I get totally stuck on what usually ends up being a minor point.

Any tips?
 

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