Prove if the following sequence is convergent $$ f(n) =\frac{6n^3 + 2n+(−4)^n}{4^n-1} $$
The sequence is dominated by $4^n$, so we divide by the dominant term, and we get $(-1)^n$ which is not a null sequence. So it is not monotonic and not convergent.
Is that right?
 A: We can rewrite the equation as:
$$f(n) = 6 \frac{n^3}{4^{n}-1} + 2 \frac{n}{4^{n}-1} +\frac{(-4)^n}{4^n -1}.$$
The first two terms will go to zero as $n\to \infty$. You can show this by methods such as l'Hopital's rule. Or just noting that an exponential will be much larger than a power function in the long run.
As for the last term, we can divide the top and bottom by $4^n$ this gives:
$$(-1)^n \frac{1}{1-\frac{1}{4^n}}.$$
As $n\to\infty$, $(1/4)^n \to 0$. Which means the fraction will go to $1/1=1$. However, since we are still alternating the sign, this sequence will ultimately alternate between $-1$ and $1$. This means that the sequence is not convergent. It is also not monotonic, since going from $-1$ to $1$, it increases and going from $1$ to $-1$, it decreases.
A: We can compare $f$ with  a smaller sequence (although as $n$ gets large the difference between the two is arbitrarily small)
$$\begin{align}\frac{6n^3 + 2n+(−4)^n}{4^n-1} > \frac{6n^3 + 2n+(−4)^n}{4^n} \\ \approx \frac{(-4)^n}{4^n}\\  =  \frac{4^n(-1)^n}{4^n} \\ = (-1)^n\end{align}$$ You are correct that $f$ is neither monotonic nor convergent.
A: Hint: prove that the two sequences $(f(2n))_n$ and $(f(2n+1))_n$ both converge, but not to the same limit.
