Is $(-3)^n + 5^n$ monotone? How can I establish if this sequence is monotone?
If it isn't, is it permanently monotone from a certain n* to infinite?
 A: Suppose
$$(-3)^{n+1} + 5^{n+1} < (-3)^n + 5^n.$$
Then $4\cdot 5^n<4\cdot (-3)^n$. Hence $5^n<(-3)^n$, a contradiction.
A: Let $a_n=(-3)^n+5^n$. Then we have $a_n=2a_{n-1}+15a_{n-2}$. So as soon as you have two successive positive terms, it is easy to prove that the sequence is increasing.
Here we use that if $a_n=A\alpha^n+B\beta^n$ then $a_n=(\alpha+\beta)a_{n-1}-\alpha\beta a_{n-2}$ which comes from the general theory of recurrence relations.
A: To show that a sequence is stictly increasing, we must show that $a_{n+1} > a_n$ for all $n$.
\begin{array}
1(-3)^{n+1}+5^{n+1} &>& (-3)^n + 5^n \\
5^{n+1} - 5^n &>& (-3)^n - (-3)^{n+1} \\
5^n (5-1) &>& (-3)^n (1+3) \\
5^n &>& (-3)^n
\end{array}
When $n$ is even then we have $5^n > 3^n$ which is clearly true. When $n$ is odd we have $5^n > -3^n$ which is clearly true since $5^n$ is positive and $-3^n$ is negative. Hence your sequence increases for all $n\ge 1.$
A: Hint:
$$\frac{(-3)^{n+1}+5^{n+1}}{(-3)^n+5^n}=\frac{5^{n+1}\left(\left(-\frac35\right)^{n+1}+1\right)}{5^n\left(\left(-\frac35\right)^n+1\right)}\xrightarrow[n\to\infty]{}5>1$$
