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How would I use the Limit Comparison Test on this problem?

$$\sum_{n=1}^{\infty}\frac{4n+5^n}{4n+8^n}$$

I'm going to let:

$$a_n=\frac{4n+5^n}{4n+8^n}$$

$$b_n=\frac{5^n}{8^n}$$

And now I should be able to test for:

$$\lim _{n\to \infty }\frac{a_n}{b_n}=L>0$$

$$\lim _{n\to \infty }\left[\frac{\left(4n+5^n\right)}{\left(4n+8^n\right)}\cdot \frac{8^n}{5^n}\right]$$

$$\lim _{n\to \infty }\frac{4n\cdot 8^n+40^n}{4n\cdot 5^n+40^n}$$

... which is infinity over infinity, so I can use L'Hospital's rule:

$$\lim _{n\to \infty }\frac{4n\cdot 8^n\cdot \ln \left(8\right)+4\cdot 8^n+40^n\cdot \ln \left(40\right)}{4n\cdot 5^n\cdot \ln \left(5\right)+4\cdot 5^n+40^n\cdot \ln \left(40\right)}$$

... which is still infinity over infinity and leaves me no closer to simplifying the fraction.


Instead of using L'Hospital's rule in step 4, am I allowed to use the dominating term effect and replace it with:

$$\lim _{n\to \infty }\frac{40^n}{40^n}=1$$

Since, for large $n$:

$$40^n>n\cdot 8^n$$

$$40^n>n\cdot 5^n$$

(Note: I'm not sure if these 2 relationships are true or not, how would we prove them?)

If so, is this the recommended method of solving this problem, or are there other ways?

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@KannappanSampath: I'm trying to learn how to solve it both ways and didn't want too much info on each question. –  Senseful Mar 22 '12 at 20:05
    
Makes sense. :-) –  user21436 Mar 22 '12 at 20:06

2 Answers 2

up vote 1 down vote accepted

As soon as you have obtained your first expression for the ratio, divide numerator and denominator by $5^n \times 8^n$.

You will get with no trouble $$\frac{1+\frac{4n}{5^n}}{1+\frac{4n}{8^n}}.$$ Now it is time to find the limit. I think that it is clear that $$\lim_{n\to\infty}\frac{4n}{5^n}=\lim_{n\to\infty}\frac{4n}{8^n}=0.$$ If we really want to give a proof of the above two facts, we can use L'Hospital's Rule, or other methods.

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$8^n \cdot 5^n$ is $40^n$, not $10^n$. My solution would be to skip L'Hospital, and multiply top and bottom by $40^{-n}$ instead, which is essentially how you justify your "dominating term effect". This is a textbook example of when L'Hospital's rule doesn't actually help.

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I updated the question to clarify where I'm getting $10^n$ from. –  Senseful Mar 22 '12 at 20:08
    
that's still not correct; $40^n/4$ is not $10^n$. –  Dustan Levenstein Mar 22 '12 at 20:09
    
Oh oops... good catch, I updated the question. –  Senseful Mar 22 '12 at 20:13

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