A class of sequences is of bounded variation Let $\left\{ a_n \right\}$ be a null sequence s.t. 
$$\sum_{n=1}^\infty \left( \frac{1}{n}\sum_{k=n}^\infty \vert\Delta a_k\vert^p\right)^\frac{1}{p}<\infty$$
for some $p>1$.
How to prove that $\left\{ a_n \right\}$ must be of bounded variation?
My attempt: It is enough to prove that
$$\vert \Delta a_n\vert \leq \left( \frac{1}{n}\sum_{k=n}^\infty \vert\Delta a_k\vert^p\right)^\frac{1}{p}, \forall n>n_0, \text{for some} \; n_0 \in \mathbb{N}.$$
I've tried proving the last inequality using Jensens' inequality, but I can't get to the desired result.
Edit:
How to find a sequence $\left\{ a_n \right\}$ of bounded variation for which $\sum_{n=1}^\infty \left( \frac{1}{n}\sum_{k=n}^\infty \vert\Delta a_k\vert^p\right)^\frac{1}{p}<\infty$ doesn't hold for any $p>1$?

The argument is:
We wish to prove $n|\Delta a_n|^p \leq \sum_{k=n}^{\infty} |\Delta a_k|^p$.
Now suppose that the right hand side does not contain at least $n$ terms of size $|\Delta a_n|^p$. Then surely $|\Delta a_k|^p \leq |\Delta a_n|^p$ eventually (i.e. of BV). If it does contain at least $n$ terms of this size, then the inequality holds, and again $a_n$ is of bounded variation as well.
