For my newest homework I have given two functions $h,h^+:\mathbb{N}\to\mathbb{R}$ with $h(n)=n^{(-1)^n}$ and $h^+(n)=h(n+1)$. I have to proove that $h^+(n) \not\in\mathcal{O}(h(n))$ with an explicit predicate formula. Here are my thoughts so far:
$h(n)=n^{(-1)^n}=\begin{cases}\frac{1}{n}, &\text{ if }n=2k+1,\\\\n, &\text{ if }n=2k.\end{cases}$ $h^+(n)=(n+1)^{(-1)^{(n+1)}}=\begin{cases}\frac{1}{n+1}, &\text{ if }n=2k,\\\\n+1, &\text{ if }n=2k+1.\end{cases}$
Assumption: $h^+(n) \not\in\mathcal{O}(h(n))\Leftrightarrow (\forall c>0\;\forall n_0\in\mathbb{N}_0\;\exists n\geq n_0)\left[|h^+(n)|>c\cdot|h(n)|\right]\;.$
The next step was to divide both cases with even and odd $n$, however the first equation drives me crazy: $\frac{1}{n_0+1}>c\cdot n_0\text{, with }n_0=2k.$
Now I fail at rearranging the upper equation to get an expression like this: $n_0>\ldots\Rightarrow n(c):=\lceil\ldots\rceil +1\;.$
How would you prove this and can you explain me why I am not able to solve this?
Thanks for any hint!