<s>[INST] Proving that $ \sum_{k=1}^\infty \frac{k^{8} + 2^{k} }{3^{k} - 2^{k}} $ converges by the comparison test I would like to prove that the following series converges:$ \sum_{k=1}^\infty \frac{k^{8} + 2^{k} }{3^{k} - 2^{k}} $ by comparing it with a series that I already know converges. One such series could be the geometric series $ \sum_{k=1}^\infty \frac{2^{k}}{3^{k}} $. Since I know that $ \sum_{k=1}^\infty \frac{2^{k}}{3^{k}} $ converges, the only thing left to prove is that
$$ L = \lim_{k \to \infty}\frac{a_k}{b_k} < +\infty ,$$where $a_k = \frac{k^{8} + 2^{k} }{3^{k} - 2^{k}} $ and $ b_k = \frac{2^{k}}{3^{k}} $. However, when I try to prove this I get the following problem:
$$ \lim_{k \to \infty}\frac{3^{k}(k^{8}+2^{k})}{2^{k}(3^{k} - 2^{k})}  $$
And I don't know how to solve this limit problem, and show that the above limit is less than $ +\infty$ so I'm kinda stuck and would appreciate any help!
 [/INST] $$\frac{k^{8} + 2^{k} }{3^{k} - 2^{k}}= \left(\frac{2}{3}\right)^{k} \cdot \frac{1+ \frac{k^8}{2^{k}}  }{1-\left( \frac{2}{3}\right)^{k}}$$
As $\frac{1+ \frac{k^8}{2^{k}}  }{1-\left( \frac{2}{3}\right)^{k}} \to 1$, then we can say, that $\exists N \in \mathbb{N}$ such that for $k>N$ holds $\frac{k^{8} + 2^{k} }{3^{k} - 2^{k}}< \frac{3}{2} $, so $$\frac{k^{8} + 2^{k} }{3^{k} - 2^{k}}< \left(\frac{2}{3}\right)^{k-1}$$
for $k>N$.
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