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?