(First of all, sry for some spelling and similar mistakes as English is not my native language.)
I tried to prove, that the sequence $2n+(-1)^n$ is monotonically increasing, that's what I assume from some values I calculated within the interval $[-10,10]$.
I used the following equation: $$a_n \leq a_{n+1} $$ $$\Leftrightarrow 0 \leq a_{n+1}-a_n$$
I equated
$a_{n+1} = 2(n+1)+(-1)^{n+1}; a_n = 2n+1(-1)^n$
$\Leftrightarrow 0 \leq 2(n+1)+(-1)^{n+1}-2n+1(-1)^n$
$\Leftrightarrow 0 \leq 2n+2+(-1)^{n+1}-2n+1(-1)^n$
$\Leftrightarrow 0 \leq 2+(-1)^{n+1}+1(-1)^n $
$\Leftrightarrow 0 \leq 2+ \underbrace{(-1)^{n+1}+1(-1)^n}_{=0}$
$\Leftrightarrow 0 \leq 2$
This would prove that $2n+(-1)^n$ is strictly monotonically increasing, which is not the case here. The function is only monotonically increasing. What went wrong?
The function has no limits as I could figure out from function-plotters, but I'm still unable to create a formal prove for that. Can you help me with that?
I would just prove that the sequence is approaching infinity:
$$2n+1(-1)^n$$
$$\underbrace{\underbrace{\underbrace{2n}_{2\cdot\infty=\infty}+1}_{\infty+1=\infty}\underbrace{(-1)^n}_{=\pm1}}_{\infty\pm1=\infty}$$ This shows that the sequence is approaching infinity. Is that enough to prove it's not limited?
Thanks in advance,
$Doesbaddel$
Best Answer
Let's try and simplify: \begin{align} a_{n+1}-a_n &=(2(n+1)+(-1)^{n+1})-(2n+(-1)^n) \\[4px] &=2n+2+(-1)^{n+1}-2n-(-1)^n \\[4px] &=2+(-1)(-1)^n-(-1)^n \\[4px] &=2-2(-1)^n \\[4px] &=2(1-(-1)^n) \\[4px] &=\begin{cases} 0 & \text{$n$ even} \\ 4 & \text{$n$ odd} \end{cases} \\[4px] &\ge0 \end{align} Indeed, $a_0=2\cdot0+(-1)^0=1$, $a_1=2\cdot1+(-1)^1=1$, $a_2=2\cdot2+(-1)^2=5$, $a_3=2\cdot3+(-1)^3=5$ and so on.
Can you spot your mistake?
Thus the sequence is not strictly increasing, but increasing nonetheless.
You can also see that $a_0=1$, $a_2=5$, $a_4=9$, and prove that $a_{2n}=4n+1$ by induction, so the sequence is unbounded.