How to evaluate limit of a sequence $\lim_{n \to \infty} \frac{2 \cdot 3^{2n - 1} - \left( -2 \right)^n}{2 \cdot 3^n - 3 \cdot 2^{2n + 1}}$

I need a help with evaluating a limit of a sequence $$\lim_{n \to \infty} \frac{2 \cdot 3^{2n - 1} - \left( -2 \right)^n}{2 \cdot 3^n - 3 \cdot 2^{2n + 1}}.$$ The problem is that $$\lim_{n \to \infty} \left( -2 \right)^n$$ does not exist. What can we even tell from this when there's a part oscilating between $$+\infty$$ and $$- \infty$$. Wolfram says it should equal to $$- \infty$$, but how to get there? The only thing I know might help is to factor out the fastest growing terms.

$$\lim_{n \to \infty} \frac{2 \cdot 3^{2n - 1} - \left( -2 \right)^n}{2 \cdot 3^n - 3 \cdot 2^{2n + 1}} = \frac{1}{3}\lim_{n \to \infty} \frac {2 \cdot 3^{2n} - \left( -2 \right)^n}{2 \cdot 3^n - 6 \cdot 2^{2n}} = \frac{1}{3} \lim_{n \to \infty} \left( \frac {3}{2} \right)^{2n} \cdot \frac {2 - \left( - \frac {2} {9} \right)^n}{2 \cdot \left( \frac{3}{4} \right)^n - 6}$$ Now we can see that $$\left( \frac {3} {4} \right)^n$$ goes to zero. What about $$\left(- \frac{2}{9} \right)^n$$? I guess that despite the fact that the values oscilate between $$+$$ and $$-$$, the overall fraction has to go to zero, hence giving $$\left| \frac{1}{3} \cdot \infty \cdot \frac{2}{-6} \right| = \left| - \frac{\infty}{9} \right| = -\infty$$ Does that make sense?

• Seems right. ${}$ May 26, 2020 at 19:17
• An alternative approach: Divide everything in sight by $9^n$. The limit of this quotient approaches $1/0$, which is bad news for the original term. May 26, 2020 at 19:20
• And if I wasn't lucky to get $\left( \frac {a} {b} \right)^n$ where $b>a$ the limit wouldn't exist right? May 26, 2020 at 19:35

Divide by $$3^{2n}$$, numerator becomes $$\frac{2}{3} - (-1)^n (\frac{2}{3})^n$$, so the second term converges to 0. In he denominator you also get term that converges to $$0$$ - another term that converges to $$0$$, so in total the expression has limit of the type $$\frac{1-0}{0-0}$$
• But $\left( - \frac{2}{3} \right)^n \neq \left( -1 \right) \cdot \left( \frac{2}{3} \right)^n$. Edit: Oh, you used the $\left( -1 \right)^2$ right? May 26, 2020 at 19:37
• @tomashauser: fixed. It doesn't matter though because the $(\frac{2}{3})^n$ term vanishes and so does $(-1)^n$. Or, if you don't like, it, take the upper (1) and lower (-1) bounds on this term and observe as they tend to 0, and the result for $(-1)^n$ follows by squeeze lemma