Suppose $f(n)\sim g(n)$ as $n\to\infty$. Is it necessarily true that \begin{equation}\frac{1}{n}\sum_{k=1}^n|f(k+1)-f(k)|\sim\frac{1}{n}\sum_{k=1}^n|g(k+1)-g(k)|\end{equation} as $n\to\infty$?
Intuitively this should be true; one would expect the average values of these differences to be asymptotically the same as the analogue for $g$, but I cannot figure out how to make this more rigorous. My idea was that, if the above is true, then it is also true that $\sum_{k=1}^n|f(k+1)-f(k)|\sim\sum_{k=1}^n|g(k+1)-g(k)|$ by multiplying through by $n$. But this doesn't really yield anything.
Not true. With $f(n) = n$ and $g(n)=n +10(-1)^n$, then $f(n)\sim g(n)$ and $$\frac{1}{n}\sum_{k=1}^n|f(k+1)-f(k)|=1$$ while $$\frac{1}{n}\sum_{k=1}^n|g(k+1)-g(k)|=\frac{1}{n}\sum_{k=1}^n\left|20(-1)^{k+1}-1\right| \rightarrow 20 $$