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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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    Makes sense. :-)2012-03-22

2 Answers 2

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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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    Oh oops... good catch, I updated the question.2012-03-22