Identifying whether the sequence $2n+(-1)^n$ is monotonically increasing/decreasing and if it is limited. (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,
$Doesbaddel$
 A: 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.
A: Hint: If $n$ is even, say $$n=2m$$ we get $$2(2m)+(-1)^{2m}=4m+1$$
If $n$ is odd, say $$n=2m+1$$ then we get $$2(2m+1)+(-1)^{2m+1}=4m+2-1=4m+1$$
in both cases.
A: For all $n$, we have
$$u_n=2n+(-1)^n\ge 2n-1$$
$$\lim_{n\to+\infty}(2n-1)=+\infty\implies \lim_{n\to+\infty}u_n=+\infty$$
$(u_n)$ has no upper bound.
A: For "strictly", you need $<$! Hence, there is nothing wrong.
You could also prove that $(a_n)_n$ is unbounded, but hamams answer is the fastet way, I guess.
